Sum divides Factorial

Question

Is there a prime number $p$ such that, $$\frac{p(p-1)}{2}\mid (p-1)! $$ or does this only hold for composites?

I have extensively tried multiple examples and even tried to prove that this only holds for a composite n and have gotten this far:

Suppose $\frac{n(n-1)}{2}\mid (n-1)!$. Then $\frac{n(n-1)}{2}\cdot k=(n-1)!$ $\Leftrightarrow k=\frac{(n-1)!}{\frac{n(n-1)}{2}}\Leftrightarrow k=\frac{2(n-1)!}{n(n-1)}\Leftrightarrow k=\frac{2(n-2)!}{n}$. Yet I am at a miss right now and am not sure where to go from here.

Help if you would.

Answer 1

Your proof is (essentially) done. $k$ must be an integer. But, if $n=p$ is prime, then when will $\frac{2(p-2)!}{p}$ be integral?

Never, as $p\nmid 2(p-2)!$ for prime $p>2$. Note that $p = 2$ is a prime that does satisfy your condition though.