Prove that $(mn)!$ is divisible by $(n!)\cdot(m!)^n$ 
Let $m$ be a positive integer and $n$ a nonnegative integer. Prove that
$$(n!)\cdot(m!)^n|(mn)!$$

I can prove it using Legendre's Formula, but I have to use the lemma that
$$ \dfrac{\displaystyle\left(\sum_{i=1}^na_i\right)!}{\displaystyle\prod_{i=1}^na_i!} \in \mathbb{N} $$
I believe that it can be proved using the lemma, since in this answer of Qiaochu Yuan he has mentioned so at the end of his answer.
Any help will be appreciated.
Thanks.

In fact this follows from Lagrange's Theorem: the wreath product $S_m\wr S_n$ is a subgroup of the symmetric group $S_{mn}$. One may view $S_m\wr S_n$ as the set of permutations which preserve the set partition $\{\{1,\cdots,m\},\{m+1,\cdots,2m\},\cdots,\{(n-1)m+1,\cdots,nm\}\}$. If we call the parts of this partition $X_1,\cdots,X_n$, then choosing an element of the stabilizer amounts to choosing a permutation $\sigma:\{1,\cdots,n\}\to\{1,\cdots,n\}$ and then for each $1\le i\le n$ choosing a bijection $X_i\to X_{\sigma(i)}$, so there are $n! m!^n$ elements in the stabilizer.
