express 1^2 * 3^2 * 5^2 * .... * (p-2)^2 (mod p) as a power of -1 .
suppose p is an odd prime.
Note that:
$\displaystyle 1\equiv{-(p-1)}(\bmod.p)$; $\displaystyle 3\equiv{-(p-3)}(\bmod.p)$;...$\displaystyle p-4\equiv{-4}(\bmod.p)$; $\displaystyle p-2\equiv{-2}(\bmod.p)$
Thus: $\displaystyle (p-1)!\equiv{(-1)^{\left(\frac{p-1}{2}\right)}\cdot{1^2\cdot{3^2}\cdot{...\cdot{(p-2)^2}}}}(\bmod.p)$
But $\displaystyle (p-1)!\equiv{-1}(\bmod.p)$ by Wilson's Theorem
Thus: $\displaystyle 1^2\cdot{3^2}\cdot{...\cdot{(p-2)^2}}\equiv{(-1)^{1-\left(\frac{p-1}{2}\right)}}(\bmod.p)$