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?
 A: We have
\begin{align*}
\frac{(sr)!}{(r!)^s}
&=\frac{(sr)!}{r!\cdot[(s-1)r]!}
  \cdot\frac{[(s-1)r]!}{r!\cdot[(s-2)r]!}\cdots\frac{r!}{r!}=
{sr\choose r}\cdot{(s-1)r\choose r}\cdots{r\choose r}\in\mathbb{N}.
\end{align*}
A: Any of the products 
$1\cdot 2 \cdot 3... \cdot r$, 
$(r+1)\cdot (r+2)\cdot (r+3)\cdot ... 2r$,
... 
$(s-1)r\cdot ((s-1)r+1) \cdot ((s-1)r+2)\cdot ...\cdot sr$
is divisible by $r!$ because dividing the products by $r!$, we simply get the binomial coefficients $\binom {kr}{r}$ , $k=1,...,s$
A: 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 :
$$
\eqalign{
(1 \cdot 2  \cdot 3  ) \ &| \ \ (1 \cdot 2  \cdot 3) \ obvious \cr
(1 \cdot 2  \cdot 3  ) \ &| \ \ (4 \cdot 5  \cdot 6) \ \ because\ \ \binom{6}{3}\ is \ integer\cr
(1 \cdot 2  \cdot 3  ) \ &| \ \ (7 \cdot 8  \cdot 9) \ \ because\ \ \binom{9}{3}\ is \ integer\cr
(1 \cdot 2  \cdot 3  ) \ &| \ \ (10 \cdot 11  \cdot 12) \ \ because\ \ \binom{12}{3}\ is \ integer \cr   
}
$$
A: You have $r$ balls each of $s$ different colors. In how many ways can you arrange these $rs$ balls in a line?
A: $S_r \times S_r \times \cdots \times S_r$ ($s$ times) is a subgroup of $S_{rs}$.
By Lagrange's theorem in group theory, its order $(r!)^s$ divides $(rs)!$, the order of $S_{rs}$.
More generally, if $n=n_1+n_2+\cdots+n_k$, then
$n_1! n_2!\cdots n_k!$ divides $n!$ because
$S_{n_1} \times S_{n_2} \times \cdots \times S_{n_k}$ is a subgroup of $S_{n}$, whose index is the multinomial coefficient
$$
\binom{n_1+n_2+\cdots+n_k}{n_1! n_2!\cdots n_k!}
$$
