How is $\left(\frac {p-1}{2}\right)! ^2 \equiv -1 \pmod p$. I was using this result for a proof ($x^2$ = -1 (mod p) ) and was stuck on this part. NOTE, this result is only when p is a prime of form 4k+1.
note: please don't delete my question without atleast informing me of the reason.
Your claim seems false:
Take $ p=7$.
$$(\frac{p-1}{2})!=3!=6$$
$$6^2=36\equiv 1 \mod 7$$