Why $(n!)^m\cdot m!$ is divisor of $(mn)!$

Question

For natural number $m, n$, i found out that $\{\{\{a_1, a_2, \cdots a_n\},\{a_{n+1}, a_{n+2}, \cdots a_{2n}\}, \cdots \{a_{mn-n+1}, \cdots a_{mn}\}\}| \{a_1,a_2, \cdots a_{mn}\}=\{1,2, \cdots mn\}\}$ has $\frac{(mn)!}{(n!)^mm!}$ elements. This would be combinatorial proof that $(n!)^mm!$ is divisor of $(mn)!$. However, I want to know the 'straight proof' using number theory. Any help?

Answer 1

There are $m$ runs of $n$ consecutive numbers, so $n!^m|(mn)!$.

Furthermore, the $k^{th}$ run (i.e. $kn+1,\dotsm (k+1)n$) leaves at least $k$ when divided by $n!$, which together multiply to $m!$.