Wilson's Theorem Application to Quadratic Residues

Question

Im a noob and trying to prove: $(m!)^2\equiv(-1)^{m+1}\textrm{(mod p)}$ using Wilson's Theorem. Although it's on Wikipedia, I want to really understand what is happening. Could someone explain how $$\begin{align*}1*2\dotsb*(p-1)=1*(p-1)*2*(p-2)\dotsb m*(p-m)\end{align*}$$ What is m? How is it possible for (p-1)! to have this expansion. Then could someone also explain the property that was used to transform this: $$(-1)^m(m!)^2\equiv-1\textrm{(mod p)}$$ into $$(m!)^2\equiv(-1)^{m+1}\textrm{(mod p)}$$ I understand that $-1^1*-1^m=-1^{m+1}$, but how did the $-1^m$ move from the left side to the right. Thanks for help! Wilson's Theorem Quadratic Residue

Answer 1

$m$ is $(p-1)/2$.

You can multiply a congruence by the same thing on both sides. Also, $(-1)^m\cdot(-1)^m=(-1)^{2m}=1$. Thus the result.