Modular arithmetic and some applications Show that if $p> 2$ is a prime number, then $p$ divides $(p-2)! - 1$. I have tried using Fermat's Theorem, but I could not solve it.

If given $p>2$: $p$ divides $(p-2)! -1$  if and only if there is an $n \in \mathbb{Z}$ for which 
$$(p-2)! -1 = np \iff (p-2)! = 1 + np$$
Modular arithmetic says an equivalent thing then:
$$(p-2)! \equiv 1 \mod p \iff (p-1)!=(p-1)(p-2)! \equiv p-1 \mod{p} $$ $$\equiv -1 \mod p$$
And this, if and only if, $p$ is a positive prime number. (Wilson's Theorem)
More clearly now:

$p>2$ is a prime number is equivalent to $p>2$ being a divisor of $(p-2)!-1$.

