Singapore Mathematical Olympiad (SMO)

First, for each $i=1,2,\dots ,\frac{p-1}{2}$, $$\frac{2i}{p}\left(\genfrac{}{}{0ex}{}{p}{2i}\right)=\frac{(p-1)(p-2)\cdots (p-(2i-1))}{(2i-1)!}\equiv \frac{(-1)(-2)\cdots (-(2i-1))}{(2i-1)!}\equiv -1(\mathrm{mod}p).$$ Hence $$\begin{array}{rl}\sum _{i=1}^{(p-1)/2}{i}^{p-2}& \equiv -\sum _{i=1}^{(p-1)/2}{i}^{p-2}\frac{2i}{p}\left(\genfrac{}{}{0ex}{}{p}{2i}\right)\\ & \equiv -\frac{2}{p}\sum _{i=1}^{(p-1)/2}{i}^{p-1}\left(\genfrac{}{}{0ex}{}{p}{2i}\right)\\ & \equiv -\frac{2}{p}\sum _{i=1}^{(p-1)/2}\left(\genfrac{}{}{0ex}{}{p}{2i}\right)(\mathrm{mod}p)\text{(by Fermat’s Little Theorem.)}\end{array}$$ The last summation counts the even-sized nonempty subsets of a $p$-element set, of which there are $2^{p-1}-1$.