Problem involving factorials (divisibility)

Question

Show that, for every $n \in \Bbb N$, the following number is natural: $$\frac {(n!)!} {{n!}^{(n-1)!}}$$. I dont't know how to prove, as I tried to find a way including combinatorics.

Answer 1

Using the multinomial coefficient approach, we have

$$(n!)!\over n!\cdot n!\cdots n!$$

and the $\cdots$ represent the product over $(n-1)!$ total terms. Since each term is $n!$, we simply sum these indices and arrive at $n\sum_{i=0}^{(n-1)!}1=n!$, and we get the multinomial coefficient

$$\binom {n!}{n,n,n,\dots,n}$$

which is always an integer.

Answer 2

$(n!)!$ is the product of $n! = n\cdot(n-1)!$ consecutive numbers. But if $a+1,a+2,\ldots,a+n$ are $n$ consecutive integers, then: $$ \frac{(a+1)(a+2)\cdot\ldots\cdot(a+n)}{n!} = \binom{a+n}{n}\in\mathbb{Z},$$ so your ratio is an integer, as the product of $(n-1)!$ integers.