Proving that $\frac{(k!)!}{k!^{(k-1)!}}$ is an integer I have to prove that:
$$\frac{(k!)!}{k!^{(k-1)!}} \in \Bbb Z$$
for any $k \geq 1, k \in \Bbb N$
Tried doing $t = k!$ which would give $$\frac{t!}{t^{t/k}}$$
But I think I just made it harder, and I have no other clue!
 A: We have
$$(n!)!=\prod_{i=1}^{n!}i=\prod_{k=1}^{(n-1)!}u_{n,k}$$where$$u_{n,k}=\prod_{i=kn-n+1}^{kn}i=(kn-n+1)\cdots(kn)=n!{kn\choose{n}}$$
hence 
$$(n!)!=(n!)^{(n-1)!}\left(\prod_{k=1}^{(n-1)!}{kn\choose n}\right)$$
A: The multinomial coefficient
$$
{n\choose k_1,k_2,\ldots, k_r},
$$
where all the variables are non-negative integers and $k_1+k_2+\cdots+k_r=n$, counts the number of ways we can partition a set of $n$ objects into subsets $A_1,A_2,\ldots,A_r$ such that $|A_i|=k_i$ for each $i$. Alternatively it is the coefficient of the term $x_1^{k_1}x_2^{k_2}\cdots x_r^{k_r}$ in the multinomial expansion of $(x_1+x_2+\cdots+x_r)^n$.
It is easy to show by induction on $r$ (using the more common binomial coefficient as the base case as well as the inductive step) that
$$
{n\choose k_1,k_2,\ldots, k_r}=\frac{n!}{k_1!k_2!\cdots k_r!}.
$$
The number in the question is the multinomial coefficient with $n=k!$, $r=(k-1)!$ and $k_i=k$ for all $i=1,2,\ldots,(k-1)!$.
Therefore it is an integer.
A: A hint:
Assume that there are $(k-1)!$ colors, and that you have $k$ balls of each color. In how many ways can you arrange them in a long line?
