prove that $(n!)^{(n-1)!}$ divides (n!)!

Question

prove that $(n!)^{(n-1)!}$ divides (n!)!

I know this question already exists but i'm looking for a purely number theory proof, no combinatorics.

My attempt: I tried to go about the concept of largest prime power that divides n! , which is given by $[\frac np]+[\frac{n}{p^2}] +......$ upto infinity (where [.] is the greatest integer function). So i tried to prove that largest power of prime p that divides (n!)! ≥ largest power of p that divides $(n!)^{(n-1)!}$, but i ended up with an ugly inequality with no idea how to proceed further.

Any help would be appreciated, cheers!

Answer 1

Pick a prime number $p$. For every $m\in \mathbb{Z}$ let $\alpha_p(m)$ be the exponent of $p$ in factorization of $m$. Then $$\alpha_p(n!) = \sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]$$ and $$\alpha_p((n!)!) = \sum_{k=1}^{+\infty}\bigg[\frac{n!}{p^k}\bigg]$$ In particular, we have $$\alpha_p\big((n!)^{(n-1)!}\big) = (n-1)!\cdot \alpha_p(n!) = (n-1)!\cdot \bigg(\sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]\bigg)$$ Since $$x\cdot [y] \leq [x\cdot y]$$ for $x,y>0$ and $x\in \mathbb{Z}$, we deduce that $$\alpha_p\big((n!)^{(n-1)!}\big) = (n-1)!\cdot \bigg(\sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]\bigg) =$$ $$= \sum_{k=1}^{+\infty}(n-1)!\cdot \bigg[\frac{n}{p^k}\bigg] \leq \sum_{k=1}^{+\infty}\bigg[(n-1)!\cdot \frac{n}{p^k}\bigg] = $$ $$= \sum_{k=1}^{+\infty}\bigg[\frac{n!}{p^k}\bigg] = \alpha_p\big((n!)!\big)$$

Answer 2

You can partition $n!$ elements into $(n-1)!$ groups of $n$ elements.

Therefore, the group $S_n \times S_n \times \cdots \times S_n$ (with $(n-1)!$ factors) is a subgroup of $S_{n!}$.

The claim follows by Lagrange's theorem of group theory.