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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!

user1176409
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Aditya_math
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    https://math.stackexchange.com/questions/3050842/need-help-understanding-a-step-in-an-induction-proof proves it by induction. – player3236 Nov 06 '20 at 07:19
  • See also https://math.stackexchange.com/questions/357032/proving-that-frackkk-1-is-an-integer all therein linked posts. – Sil Sep 21 '21 at 07:16

2 Answers2

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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.

lhf
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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 $$\alpha_p\big((n!)^{(n-1)!}\big) = (n-1)!\cdot \bigg(\sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]\bigg) =$$ $$= \sum_{k=1}^{+\infty}(n-1)!\cdot \bigg[\frac{n}{p^k}\bigg] \leq \sum_{k=1}^{+\infty}\bigg[(n-1)!\cdot \frac{n}{p^k}\bigg] = $$ $$= \sum_{k=1}^{+\infty}\bigg[\frac{n!}{p^k}\bigg] = \alpha_p\big((n!)!\big)$$

Slup
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    could you please tell me how to prove ⋅[]≤[⋅]? thanks ! – Aditya_math Nov 06 '20 at 07:40
  • Sorry. I was confused for a moment. This inequality holds for $x\in \mathbb{Z}$. That was a gap in the argument. I fixed it thanks to your comment. – Slup Nov 06 '20 at 07:46
  • For the proof note that $x\cdot [y]$ is an integer (because $x\in \mathbb{Z}$) and $x\cdot [y] \leq x\cdot y$ (because $x,y>0$). Hence $x\cdot [y] \leq [x\cdot y]$ by definition. – Slup Nov 06 '20 at 07:48
  • i tried it this way, [y] = y -{y}, x[y]=xy-x{y} , [x.y] = xy -[x{y}] and [x{y}]<=x{y} – Aditya_math Nov 06 '20 at 07:56
  • also, thank you so much for this answer, appreciate it! – Aditya_math Nov 06 '20 at 07:58