Show that $(r!)^2 ≡ (−1)^{r−1} \pmod p,\ r = (p-1)/2$ I need to prove that if p is an odd prime and 
$r = (p-1)/2$ then 
$(r!)^2 ≡ (−1)^{r−1} \pmod p$
I think it has something to do with gauss's lemma 
https://en.wikipedia.org/wiki/Gauss%27s_lemma_(number_theory)
but I tried a lot and couldn't find a way to break it . 
any help or hint will be appreciated.

Hint: We have $p-1\equiv -1\pmod{p}$, $p-2\equiv -2\pmod{p}$,  $p-3\equiv -3\pmod{p}$, and so on up to $\frac{p+1}{2}\equiv -\frac{p-1}{2}\pmod{p}$. 
It follows that 
$$(p-1)!\equiv (-1)^{(p-1)/2}\left(\left(\frac{p-1}{2}\right)!\right)^2\pmod{p}.$$
Now use Wilson's Theorem.
