TITLE: Combinatorial proof that $\frac{({10!})!}{{10!}^{9!}}$ is an integer
QUESTION [8 upvotes]: I need help to prove that the quantity of this division : 
$\dfrac{({10!})!}{{10!}^{9!}}$
is an integer number, using combinatorial proof

REPLY [17 votes]: You have $10!$ students. You want to divide them into groups of $10$ and line of the groups; in how many different ways can you do this?
You can line up the students in $(10!)!$ different orders. Now imagine that you count them off by tens and mark a line on the ground between groups of $10$; you have $\frac{10!}{10}=9!$ groups of $10$. Since all you care about is the order of the groups, you can allow the students in each group of $10$ to rearrange themselves within the group as they please, and you’ll still have the same lineup of groups. Each group can rearrange itself in $10!$ different orders, and there are $9!$ groups, so there are $10!^{9!}$ different lineups of students that produce the same lineup of groups. The number of lineups of groups is therefore $\frac{(10!)!}{10!^{9!}}$, which of course must be an integer.
You can replace $10$ by $n$ and $9$ by $n-1$ and repeat the argument to show that $\dfrac{(n!)!}{n!^{(n-1)!}}$ is an integer.