TITLE: Proving that $\frac{(k!)!}{k!^{(k-1)!}}$ is an integer
QUESTION [15 upvotes]: 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!

REPLY [6 votes]: 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 
$$(n!)!=(n!)^{(n-1)!}\left(\prod_{k=1}^{(n-1)!}{kn\choose n}\right)$$