Question 124
Let $p$ be an odd prime. Prove that
$$
\sum_{k=1}^{p-1} k^{2p-1} \equiv \frac{p(p+1)}{2} \quad\left(\bmod p^{2}\right)
$$
[Note that $a \equiv b(\bmod m)$ means that $a-b$ is divisible by $m$.]
$$
\sum_{k=1}^{p-1} k^{2p-1} \equiv \frac{p(p+1)}{2} \quad\left(\bmod p^{2}\right)
$$
[Note that $a \equiv b(\bmod m)$ means that $a-b$ is divisible by $m$.]