For which $n$ does $n\mid 1^n+2^n+\cdots+n^n$? Find all the natural numbers $n$ such that
$$n\mid 1^n+2^n+\cdots+n^n.$$
We know through Faulhaber's formula, that 
$$\sum_{k=1}^{n}k^n=\frac1{n+1}\sum_{k=0}^n\binom{n+1}{k}B_k n^{n+1-k},$$
where $B_k$ is a Bernoulli number. I checked few dozen values of $n$ and it seems that only odd numbers are solutions. Any ideas on how to proceed from here?

For odd $n$ we can telescope as noted by others and observe that $n\mid1^n+\cdots+n^n$.
For even $n$, let $2^k$ be the highest power of $2$ dividing $n$. By Euler's theorem, we have $a^n\equiv1\pmod{2^k}$ if $a$ is odd. Because $a^n\equiv0\pmod{2^k}$ if $a$ is even, we have
$$1^n+\cdots+n^n\equiv\frac n2\cdot1\not\equiv0\pmod{2^k}$$
(there are exactly $\frac n2$ odd terms in the sum), meaning that $n$ cannot divide $1^n+\cdots+ n^n$.
