$(p-1)!=1\cdot2\cdots\frac{p-1}{2}\cdot\frac{p+1}{2}\cdots(p-2)(p-1)$ 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 $\left[\left(\frac{p-1}{2}\right)!\right]^{2}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}$.