Prove that $(p-2)! \equiv 1\mod p$ without using Wilson's Theorem. This question is from a previous exam paper that I am using to revise. As per the rules of the School of Maths and Stats in my university they are not allowed to give out solutions to previous exams.
I know there is some "identity formula" that I could use, but I don't know what it is.
 A: Impossible - it's trivially equivalent to Wilson - put $\,n=(p-2)!\,$ below. 
$$\bmod p\!:\ \,  n\equiv 1\iff (p\!-\!1)\,n\equiv -1\qquad\qquad $$
So the congruences are just negations $\!\bmod p\,$  of each other. Any proof of $\,(p\!-\!2)!\equiv 1\pmod{\!p}$ immediately yields a proof of Wilson's Theorem by simply multiplying by: $\,p\!-\!1\equiv -1$.
More generally Wilson's Theorem is equivalent to the  Wilson reflection formula below 
$$\quad (p\!-\!1\!-\!k)!\equiv\frac{(-1)^{k+1}}{k!}\!\!\!\pmod{\!p},\ \ 0\le k < p\,\ {\rm prime}$$
OP is the special case $\ k = 1$.
A: As mentioned in the comments, we can consider $\mathbb{Z}_{p}^{*} = \{1,2,\dots, p-1\}$. Define the set $S = \{ x \in \mathbb{Z}_{p}^{*} : x^{2} \not \equiv 1 \pmod{p} \} = \mathbb{Z}_{p}^{*} \setminus \{1,p-1\}$. As a result, $\prod_{x \in S} x \equiv 1 \pmod{p}$, because $x$ and $x^{-1}$ are distinct elements in $S$. Hence, 
$$
(p-1)! = (p-1)(p-2)!= \prod_{x \in \mathbb{Z}_{p}^{*}} x = 1 \cdot (p-1) \cdot \prod_{x \in S}x \equiv (p-1)\prod_{x \in S} x\pmod{p} 
$$
This implies that:
$$
(p-1)(p-2)! \equiv (p-1)\prod_{x\in S} x  \pmod{p} \implies (p-2)! \equiv \prod_{x \in S}x \equiv 1\pmod{p}
$$
Note that we may multiply by $(p-1) \equiv -1 \pmod{p}$ on both sides to make the deduction. Also, we relied heavily on the fact that $\mathbb{Z}_{p}$ is a field, so $\mathbb{Z}_{p}^{*}$ is a group under multiplication mod $p$. 
