If $p$ is a prime, $p > 2$, prove that $1^2 3^2\dots (p-2)^2 \equiv (-1)^\frac{p+1}{2}\pmod{p}$

Question

If $p$ is a prime, $p > 2$, prove that $1^2 3^2\dots (p-2)^2 \equiv (-1)^\frac{p+1}{2}\pmod{p}$

I suppose that we could use Wilson's theorem:

$$(p-1)! \equiv -1 \pmod{p}$$

I am stuck. I would really appreciate help.

Answer 1

Starting with $(p-1)!\equiv-1$: $$1\cdot3\cdots(p-2)\;\;\cdot\;\;2\cdot4\cdots(p-1)=(-1)$$ multiply each even number on the left by $-1$ to obtain $$1\cdot3\cdots(p-2)\;\;\cdot\;\;(-2)\cdot(-4)\cdots(-p+1)=(-1)\cdot(-1)^{(p-1)/2}$$ then replace each factor $(-2n)$ by $(p-2n)$ to have: $$1\cdot3\cdots(p-2)\;\;\cdot\;\;(p-2)\cdot(p-4)\cdots3\cdot1=(-1)\cdot(-1)^{(p-1)/2}$$ as desired.