I need to prove the following:
$$\sum_{n = 1}^{p - 1} n^{p - 1} \equiv (p - 1)! + p \pmod {p^2}$$
...with $p$ being an odd prime number. The statement is obviously true for$\pmod p$ because left-hand side is congruent to $-1 \pmod p$ by Fermat's little theorem, and the right-hand side also turns out to be congruent to $-1 \pmod p$ by Wilson's theorem. Now, I am not sure how to make a jump from$\pmod p$ to$\pmod {p^2}$, if that is even possible. Maybe the sum on the left could somehow be modified using the existence of the primitive root$\pmod {p^2}$.
EDIT: Elementary solution can be found here: https://mathoverflow.net/a/319824/134054