A number-theoretic proof that $(n^2)!/(n!)^{n+1}$ is an integer

Question

I have seen a number of combinatorial proofs for this statement. For instance,

Quotient of factorials

However, I am wondering whether there is a purely number-theoretic proof.

Answer 1

Here’s such a proof: note that $(n^2)!=n^n n! \prod_{k=0}^{n-1}{\left[\prod_{l=kn+1}^{kn+(n-1)}{l}\right]}$.

Thus $\frac{(n^2)!}{n!^{n+1}}=\prod_{k=0}^{n-1}{\binom{kn+(n-1)}{n-1}}$ is an integer.

Answer 2

This may be a guide on solving the problem but I yet been able to prove it with rigorous.

We can probably prove by letting $p_k$ be the $k$th prime. We can let $n!=p_1^{a_1}p_2^{a_2}\cdots$ and $(n^2)!=p_1^{b_1}p_2^{b_2}\cdots$
We can notice that:
$a_i=\sum_{s=1}^{\infty}\lfloor\frac{n}{p_i^s}\rfloor$
$b_i=\sum_{s=1}^{\infty}\lfloor\frac{n^2}{p_i^s}\rfloor$
Hence, we only have to prove $b_i\ge(n+1)a_i$ for all $i$ such that $p_i\le n$ as it is trivial for any $p_i>n$ to not be present in the denominator.
$\Rightarrow \sum_{s=1}^{\infty}\lfloor\frac{n^2}{p_i^s}\rfloor\ge(n+1)\sum_{s=1}^{\infty}\lfloor\frac{n}{p_i^s}\rfloor$
If we are able to prove $\lfloor\frac{n^2}{p_i^s}\rfloor\ge(n+1)\lfloor\frac{n}{p_i^s}\rfloor$, then it shall hold.
I am unable to prove it yet but it can be seen easily by using cases or solved graphically.