<s>[INST] How one can show $(r!)^s$ divides $(rs)!$? I would like if anybody have suggestions to prove: 
if $n = rs$
with: $r > 0$ and $s > 0$
then $(r!)^s \mid n!$
suggestions?
 [/INST] Let illustrate it with an example. Say to prove that
$$
3!^4 | 12!
$$
We may analyze the products as follows:
$$
\eqalign
{
1 \cdot 2  \cdot 3  \cdot \ \ \ &\   \ \ \ \  1 \cdot 2  \cdot 3  \cdot \cr
1 \cdot 2  \cdot 3  \cdot \ \ \ &\   \ \ \ \  4 \cdot 5  \cdot 6  \cdot \cr
1 \cdot 2  \cdot 3  \cdot \ \ \ &|   \ \ \ \  7 \cdot 8  \cdot 9  \cdot \cr
1 \cdot 2  \cdot 3   \cdot      \ \ \ &\   \ \ \ \  10 \cdot 11  \cdot 12   \cr
}
$$
Now you should observe that the following hold :
$$
\eqalign{
(1 \cdot 2  \cdot 3  ) \ &| \ \ (1 \cdot 2  \cdot 3) \ obvious \cr
(1 \cdot 2  \cdot 3  ) \ &| \ \ (4 \cdot 5  \cdot 6) \ \ because\ \ \binom{6}{3}\ is \ integer\cr
(1 \cdot 2  \cdot 3  ) \ &| \ \ (7 \cdot 8  \cdot 9) \ \ because\ \ \binom{9}{3}\ is \ integer\cr
(1 \cdot 2  \cdot 3  ) \ &| \ \ (10 \cdot 11  \cdot 12) \ \ because\ \ \binom{12}{3}\ is \ integer \cr   
}
$$
</s>