Why does $(\frac{p-1}{2}!)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$? Assume $p$ is prime and $p\ge 3$.
Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying $2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$, but I can't figure out any more than that.

$$(p-1)!=1\cdot2\cdots\frac{p-1}{2}\cdot\frac{p+1}{2}\cdots(p-2)(p-1)$$ We have the congruences
$$\begin{align*}
p-1&\equiv -1\pmod p\\
p-2&\equiv -2\pmod p\\
&\vdots\\
\frac{p+1}{2}&\equiv -\frac{p-1}{2}\pmod{p}\end{align*}$$
Rearranging the factors produces
$$(p-1)!\equiv 1\cdot(-1)\cdot2\cdot(-2)\cdots\frac{p-1}{2}\cdot\left(-\frac{p-1}{2}\right)\pmod p.$$
$\therefore (p-1)!\equiv (-1)^{\frac{p-1}{2}}\left(1\cdot2\cdots\frac{p-1}{2}\right)^{2}\pmod p.$
From Wilson's theorem, $(p-1)!\equiv -1\pmod{p}$
Thus $$-1\equiv (-1)^{\frac{p-1}{2}}\left[\left(\frac{p-1}{2}\right)!\right]^{2}\pmod{p}$$
It follows that $\left[\left(\frac{p-1}{2}\right)!\right]^{2}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}$.
