# How to prove that $\frac{(mn)!}{m!(n!)^m}$ is an integer?

$\forall m,n\in\mathbb Z$ , $m\ge1$ and $n\ge1$ how to prove that $$\frac{(mn)!}{m!(n!)^m}$$ is an integer?

There is a nice subgroup of the Symmetric group $S_{mn}$ whose order is $m!(n!)^{m}$. If you can see this, you can apply Lagrange's theorem. –  Geoff Robinson Feb 23 '13 at 18:54