1. Find all prime numbers $p$ such that
$$
1+2^{p}+3^{p}+\cdots+p^{p}
$$

is a prime number.
1. When $p=2$, $1+2^{2}=5$ is a prime number.

When $p>2$, $p$ is odd, then
$$
\begin{array}{l}
1+2^{p}+3^{p}+\cdots+p^{p} \\
=\left(1^{p}+(p-1)^{p}\right)+\left(2^{p}+(p-2)^{p}\right)+\cdots+ \\
\quad\left(\left(\frac{p-1}{2}\right)^{p}+\left(\frac{p+1}{2}\right)^{p}\right)+p^{p}
\end{array}
$$

can be divided by $p$, because each number in the parentheses can be divided by $p$.
$$
\begin{array}{l}
\text { By } k^{p} \equiv k^{p}(\bmod p), \\
p-k \equiv-k(\bmod p), \\
(p-k)^{p} \equiv(-k)^{p} \equiv-k^{p}(\bmod p),
\end{array}
$$

we get $k^{p}+(p-k)^{p} \equiv k^{p}-k^{p}=0(\bmod p)$. 
This shows that for every odd prime $p$,
$$
1+2^{p}+3^{p}+\cdots+p^{p}
$$

is a composite number.
Therefore, the only solution is 5 when $p=2$.