Prove that $(n!)!$ is a multiple of $(n!)^{(n-1)!}$

Question

Problem: Prove that $(n!)!$ is a multiple of $(n!)^{(n-1)!}$

So we have to show $n!^{(n-1)!}|(n!)!$ or show $n!^{(n-1)!-1}|(n!-1)!.$

This problem hasn't been solved in AOPS too.

Answer 1

We know that the $a\times (a+1)\times\dots \times (a+n)$ is diviosible by $n!$ (proofs).

Now, $(n!)!$ is the product $1\times 2\times \dots \times (n!)$. Consider the $\frac {n!}n$ segments, namely $\{1,2,3,\dots n\}$, $\{n+1,n+2, \dots ,2n\}$, $\{2n+1,2n+2,\dots ,3n\}$, all the way upto $n!$. The product of any of these $\frac {n!}n$ factors are all multiples of $n!$. This proves that $(n!)!$ is a multiple of $(n!)^{(n-1)!}$.

Does that help?

Answer 2

The multinomial coefficient $\binom{n!}{\underbrace{n,n,\dots,n}_{(n-1)! times}}$ is equal to $\frac{(n!)!}{n!^{(n-1)!}}$, and is an integer due to it's combinatorial characterization.

Alternative: Use the formula $v_p(n!) = \frac{n-s_p(n)}{p-1}$ where $s_p(n)$ is sum of digits of $p$ in base $n$.

It follows $v_p((n!)!) = \frac{n!-s_p(n!)}{p-1}$ whereas $v_p((n!)^{(n-1)!})= \frac{(n-1)!(n - s_p(n) )}{p-1} = \frac{n! - (n-1)!s_p(n)}{p-1}$, so we must prove $s_p(n!) \leq (n-1)! s_p(n)$.

In fact we have $s_p(ab) \leq as_p(b)$ and the proof is simple by induction on $a$, using $s_p(x+y) = s_p(x) + s_p(y) - c_p(x,y)$ where $c_p(x,y)$ is the number of carries when adding $x$ and $y$ in base $p$.