Prove that $p^n \nmid ((p - 1)n!)$ for all primes $p$

Question

Prove that $p^n \nmid ((p - 1)n!)$ for all primes $p$ .

First I am thinking maybe modular arithmetic will help (although I am not sure) , and I don't know a quick and a general proof of this . Can anyone help ?

I think the question should be: "Prove that $p^n\nmid \big((p-1)n\big)!$." The problem as stated seems quite weak. — Batominovski, Aug 01 '20 at 13:08
That can be a possibility , but the question is actually what I have wrote . — Anonymous, Aug 01 '20 at 13:09
My reason for doubting is: since $\gcd(p,p-1)=1$, the claim is equivalent to $p^n\nmid n!$. This is true, but very weak. — Batominovski, Aug 01 '20 at 13:11
For $p=2$ and $2^n$ see this post. Generalize it to $p$ and $p^n$. — Dietrich Burde, Aug 01 '20 at 13:29

score 4 · Accepted Answer · answered Aug 01 '20 at 13:34

It is well-known that the maximal $k$ with $p^k\mid m!$ is given by $$ k=\left\lfloor\frac mp\right\rfloor +\left\lfloor\frac m{p^2}\right\rfloor +\left\lfloor\frac m{p^3}\right\rfloor+\cdots<\frac mp+\frac m{p^2}+\frac m{p^3}+\cdots=\frac m{p-1}$$ When $m=(p-1)n$, this implies $k<n$, as desired.

Prove that $p^n \nmid ((p - 1)n!)$ for all primes $p$

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