Quite standard for a TST P2, especially for Russia. By FLT we can let $i^{p-1}=pa_i+1$ and we have to prove that $\sum pa_i \equiv (p-1)!+1 \pmod {p^2} $. Consider $((p-1)!)^{p-1}=\prod i^{p-1}=\prod (pa_i+1) \equiv \sum pa_i+1 \pmod {p^2}$, so we need $((p-1)!)^{p-1} \equiv (p-1)!+2 \pmod {p^2}$, which is quite direct by letting $(p-1)!=Mp-1$ and expanding the brackets modulo $p^2$.