For $n,k \in {\mathbb{Z}}^{+}$ (excluding $n=1$), does $\frac{(n+k)!}{n!}$ ever equal $n!$ While investigating an integer sequence, I came across the following two OEIS entries:


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*A094331: Least k such that n! < (n+1)(n+2)(n+3)...(n+k).

*A075357: a(n) = smallest k such that (n+1)(n+2)...(n+k) is just >= n!.
The generating rule for both of these sequences is basically identical, except A094331 uses $\lt$ and A075357 uses $\le$. This made me curious whether both sequences are actually identical (and I'm not the first; David Wasserman commented the same thing when he was adding more terms to A075357.) For the purposes of the OEIS, the sequences are technically different, because A094331(1) = 1 and A075357(1) = 0 (i.e. for $n=1$ and $k=0$, $\frac{(n+k)!}{n!} = n!$.)
In approaching this problem, I first tried computational brute forcing. For values of $n$ from 2 to 1000000, $n \ne k$. However, this approach is obviously limited. Since my ability in number theory is very weak, I was wondering if anyone with a greater knowledge of number theory may be able provide a definitive answer to this question.

With the exception of $N=1$, $N!$ is never a square. This is because, by Bertrand's Postulate, there is always a prime between $N$ and $\lfloor N/2\rfloor$ (to be precise, a prime $p$ satisfying $\lfloor N/2\rfloor\lt p\le N$), and such a prime can only divide $N!$ once. So if we take $N=n+k$ with positive integers $n$ and $k$, we have $N\gt1$ and so $N!=(n+k)!$ is not a square.  In particular, it cannot be the case that $(n+k)!=(n!)^2$. Thus it is never the case that $(n+k)!/n!=n!$.
It might be of interest to see if there is a proof that $N!$ is never the square of a factorial that doesn't rely on Bertrand's Postulate.
