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

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!

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.

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

• could you please tell me how to prove 𝑥⋅[𝑦]≤[𝑥⋅𝑦]? thanks ! Commented Nov 6, 2020 at 7:40
• 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
Commented Nov 6, 2020 at 7:46
• 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
Commented Nov 6, 2020 at 7:48
• i tried it this way, [y] = y -{y}, x[y]=xy-x{y} , [x.y] = xy -[x{y}] and [x{y}]<=x{y} Commented Nov 6, 2020 at 7:56
• also, thank you so much for this answer, appreciate it! Commented Nov 6, 2020 at 7:58