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You may play with this method and find formulae for many coefficients. To be more concrete, below I prove using it that the coefficient of $(z_p/z_q)^c$ equals $-(N-c) (N-2)!$ for all $c=1,2,\dots,N-1$.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term.

Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$, as I said in the beginning. (This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.)

It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and after careful but straightforward cancellation we get the above answer.

You may play with this method and find formulae for many coefficients. To be more concrete, below I prove using it that the coefficient of $(z_p/z_q)^c$ equals $-(N-c) (N-2)!$ and the coefficient of $z_1^c(z_2\dots z_{c+1})^{-1} $ equals $(-1)^c c! (N-c)!$ for all $c=1,2,\dots,N-1$.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term. If $c_1=c>0$, $c_{N}=c_{N-1}=\dots =c_{N-c+1}=-1$, the same choice of $\lambda$'s gives you unique non-zero value, that yields the aforementioned answer $(-1)^c c! (N-c)! $.

Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$, as I said in the beginning. (This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.)

It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and after careful but straightforward cancellation we get the above answer.

You may play with this method and find formulae for many coefficients.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term.

Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$. This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.

It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and we get the answer.

You may play with this method and find formulae for many coefficients. To be more concrete, below I prove using it that the coefficient of $(z_p/z_q)^c$ equals $-(N-c) (N-2)!$ for all $c=1,2,\dots,N-1$.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term.

Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$, as I said in the beginning. (This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.)

It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and after careful but straightforward cancellation we get the above answer.

You may play with this method and find formulae for many coefficients.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$.

What happens if, say, $c_1=c>0$, $c_2=0$? This includes, in particular, the cases mentioned in the answer of Gjergji Zaimi. It makes sense to change $\lambda$'s [caution: I do not insist that the $\lambda$'s I suggest here are most clever, you are welcomed to play around]: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $c_{N}<0$, this yields $z_{N}<N-1$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. So, we have about $2c$ possibilities of $z_N$ and $z_1$, and other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{z_N,z_N+1\}$. The corresponding summands in the CN formula do not depend on the permutation and we find the answer as a sum of about $2c$ summands.

You may play with this method and find formulae for many coefficients.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term.

Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$. This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.

It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and we get the answer.