Solution
= Solution
For each $1\le j\le p-1$, <Fermat's little theorem> gives $j^p\equiv j\pmod p$. Since $p$ is odd,
$$
\boxed{\sum_{j=1}^{p-1}j^p\equiv\sum_{j=1}^{p-1}j=\frac{p(p-1)}2\equiv0\pmod p}.
$$
= Solution
For each $1\le j\le p-1$, <Fermat's little theorem> gives $j^p\equiv j\pmod p$. Since $p$ is odd,
$$
\boxed{\sum_{j=1}^{p-1}j^p\equiv\sum_{j=1}^{p-1}j=\frac{p(p-1)}2\equiv0\pmod p}.
$$