Let $p$ be an odd prime. Prove that
$$ 2 (p− 3)! \equiv −1 \pmod p $$
We start with Wilson's Theorem
$$ (p-1)! \equiv -1 \pmod p $$
That is
$$ (p-1) \times (p-2) \times (p-3)! \equiv -1 \pmod p $$
We note that $(p-1) \equiv -1 \pmod p$.
We also note that $(p-2) \equiv -2 \pmod p$, valid since an odd prime is larger than 2.
$$ (-1) \times (-2) \times (p-3)! \equiv -1 \pmod p $$
Simplifying gives us the desired result
$$ 2 (p− 3)! \equiv −1 \pmod p $$