Prove that $(n!)!$ divisible by $(n!)^{(n-1)!}$

Question

I was trying to think of a situation and use combinatorics to solve the problem. Any other arithmetic solution is also appreciated.

Answer 1

Rewrite $P=\frac{(n!)!}{(n!)^{(n-1)!}}$ as $\frac{(n!)!}{(n!)(n!) \cdots (n!)}$

Number of $n!$s in the denominator is $(n-1)!$

Now, $(n-1)! \times n = n!$. $P$ is the number of ways of arranging $n!$ things in which $n$ things are of type 1, other $n$ things are of type 2 and so on. Hence P is an integer.

Using a different combinatorial interpretation we can even say that,

$$\frac{(n!)!}{(n!)^{(n-1)!}(a_1)!(a_2)!....\cdots (a_r)!}$$ is also an integer where $a_i$ are non-negative integers and $a_1+a_2+a_3 \cdots + a_r=(n-1)!$. Try to figure out how you will prove this.