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Whether or not the same proof works for $d>2$, depends on the degenerations of rational normal curves of degree $d$ and you probably know more about these than I do. Anyway, a smooth curve in a smooth total space is a local complete intersection, so the question here is whether this remains true for all degenerations. If it does then $V_{d,n+1}$ is a local complete intersection in $V_{d,n}\times \mathbb P^d$ and then by the above argument itself is CM if the previous one was CM.

I think this definitely works for $d=2$ and possibly for other $d$'s, but as you can tell I did not work out all the details. At least this might give you some ideas on how one might go about proving something like this. I would add that if this does not turn out to be normal, then perhaps this is not the "right" compactification to consider. I could imagine that you demand that degenerations be  stable, i.e., consider the pointed curve degenerating to a pointed degenerate curve that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above proof would essentially prove that it is normal. Of course, this may be what you're doing in general or it does not work for some reason, I am not familiar with the relevant results, so this is just the first thing I would try.

The same proof does not work for $d>2$. Although a smooth curve in a smooth total space is a local complete intersection, as pointed out by mdeland in the remarks this does not remain true for all degenerations. (I should have realized this as this is a famous example...) At the same time I feel that one might still be able to do something like this. After all we do not need the individual curves to be lci or even S_2. (If we did, this example would lead to non-normality).

It seems this definitely works for $d=2$ but not for other $d$'s. However, at least this might give you some ideas on how one might go about proving something like this. I would add that if this does not turn out to be normal, then perhaps this is not the "right" compactification to consider. At the least you should assume that your degenrate curves have no embedded points, but I could imagine that the best to do is to demand that degenerations be  stable, i.e., consider the pointed curve degenerating to a pointed degenerate curve that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above proof would essentially prove that it is normal. Of course, this may be what you're doing in general or it does not work for some reason, I am not familiar with the relevant results, so this is just the first thing I would try.

I think this definitely works for $d=2$ and possibly for other $d$'s, but I did not work out all the details.

The main idea is to consider the forgetful morphism $(\mathbb P^d)^{n+1}\to (\mathbb P^d)^n$. This restricts to a forgetful morphism $\pi:V_{d,n+1}\to V_{d,n}$. This is kind of like the map $\overline{\mathscr M}_{g,n+1}\to \overline{\mathscr M}_{g,n}$. The general fiber of $\pi$ is a $\mathbb P^1$ (the original degree $d$ rational normal curve).

Perhaps this gives you some ideas on how one might go about proving this. I would add that if this does not turn out to be normal, then perhaps this is not the "right" compactification to consider. I could imagine that you demand that degenerations be stable, i.e., consider the pointed curve degenerating to a pointed degenerate curve that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above proof would essentially prove that it is normal. Of course, this may be what you're doing in general or it does not work for some reason, I am not familiar with the relevant results, so this is just the first thing I would try.

Consider the forgetful morphism $(\mathbb P^d)^{n+1}\to (\mathbb P^d)^n$. This restricts to a forgetful morphism $\pi:V_{d,n+1}\to V_{d,n}$. This is kind of like the map $\overline{\mathscr M}_{g,n+1}\to \overline{\mathscr M}_{g,n}$. The general fiber of $\pi$ is a $\mathbb P^1$ (the original degree $d$ rational normal curve).

I think this definitely works for $d=2$ and possibly for other $d$'s, but as you can tell I did not work out all the details. At least this might give you some ideas on how one might go about proving something like this. I would add that if this does not turn out to be normal, then perhaps this is not the "right" compactification to consider. I could imagine that you demand that degenerations be stable, i.e., consider the pointed curve degenerating to a pointed degenerate curve that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above proof would essentially prove that it is normal. Of course, this may be what you're doing in general or it does not work for some reason, I am not familiar with the relevant results, so this is just the first thing I would try.

here is an idea that might help you analyze the situation. In fact, I think this definitely works for $d=2$ and possibly for other $d$'s, but I did not work out all the details.

For $n=d+3$ it is easy since $V_{d,n}=(\mathbb P^d)^n$. So it is normal (more of course). In particular it is both $R_1$ and $S_2$, in fact it is Cohen-Macaulay (CM).

Whether or not the same proof works for $d>2$, depends on the degenerations of rational normal curves of degree $d$ and you probably know more about these than I do. Anyway, a smooth curve is a local complete intersection, so the question here is whether this remains true for all degenerations. If it does then $V_{d,n+1}$ is a local complete intersection in $V_{d,n}\times \mathbb P^d$ and then by the above argument itself is CM if the previous one was CM.

I think this definitely works for $d=2$ and possibly for other $d$'s, but I did not work out all the details.

For $n=d+3$ it is easy since $V_{d,n}=(\mathbb P^d)^n$. So it is normal (actually smooth of course). In particular it is both $R_1$ and $S_2$, in fact it is Cohen-Macaulay (CM).

Whether or not the same proof works for $d>2$, depends on the degenerations of rational normal curves of degree $d$ and you probably know more about these than I do. Anyway, a smooth curve in a smooth total space is a local complete intersection, so the question here is whether this remains true for all degenerations. If it does then $V_{d,n+1}$ is a local complete intersection in $V_{d,n}\times \mathbb P^d$ and then by the above argument itself is CM if the previous one was CM.