How to prove that $(p-1)^2$ $\mid$ $(p-1)!$ when $p$ is a prime number and $p>5$?

Question

I say that $p-1$ $\mid$ $(p-1)!$ then I want to prove that $p-1$ $\mid$ $(p-2)!$.

I started by saying that $p-1$ is an even number so $2\mid (p-1)$ and that means that $\frac{p-1}{2}$ is an integer.

Then I just know that $\frac{p-1}{2}<p-1$ but I don't know what to do next.

Thanks in advance !

Answer 1

Since $(p-2)>\frac{p-1}{2}>2$ and $\frac{p-1}{2}\cdot 2=(p-1)$, $(p-1)$ divides: $$ 1\cdot\color{red}{2}\cdot 3\cdot\ldots\cdot\color{red}{\frac{p-1}{2}}\cdot\frac{p+1}{2}\cdot\ldots\cdot(p-2)=(p-2)!.$$

Answer 2

In fact, if $m (m \ge 6)$ is an composite number, we have $(m - 1)! \equiv 0 \pmod m$.

Here, $m = p - 1$.