Let $p$ be prime. Prove that
$$ (p− 2)! \equiv 1 \pmod p $$
We start with Wilson's Theorem
$$ (p-1)! \equiv -1 \pmod p $$
That is
$$ (p-1) \times (p-2)! \equiv -1 \pmod p $$
Now, $(p-1) \equiv -1 \pmod p$, so
$$ (-1) \times (p-2)! \equiv -1 \pmod p $$
Multiplying by (-1), gives us the desired result
$$ (p-2)! \equiv 1 \pmod p $$