TITLE: How to prove that $\frac{(mn)!}{m!(n!)^m}$ is an integer?
QUESTION [18 upvotes]: $\forall m,n\in\mathbb Z$ , $m\ge1$ and $n\ge1$ how to prove that  $$\frac{(mn)!}{m!(n!)^m}$$ is an integer?
Thanks in advance.

REPLY [10 votes]: OK, since explicit solutions have been given, I will give the group-theoretic approach explicitly. The symmetric group $S_{nm}$ has a subgroup $S_{n} \wr S_{m}.$
This is a semidirect product of a base group $B$ with the symmetric group $S_{m}.$ The base group $B$ is itself a  direct product of $m$ copies of $S_{n},$ and these factors are permuted around by the $S_{m}$ which acts on $B.$  The group $S_{n} \wr S_{m}$ therefore has order $m!(n!)^{m},$ which divides $(mn)!$ by Lagrange's theorem.