TITLE: Show that $\sum\limits_{i=1}^{p} i^{p}$ is divisible by $p$ for all primes $p > 2$.
QUESTION [5 upvotes]: Show that
   $$
 \sum\limits_{i=1}^{p} i^{p}
 $$
   is divisible by $p$ for all primes $p > 2$.

I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to do it.

REPLY [3 votes]: And yes, it can be solved using LFT as well, since 
$$i^p\equiv i \pmod{p}$$
thus
$$\sum\limits_{i=1}^{p}i^p \equiv \sum\limits_{i=1}^{p}i \pmod{p} \tag{1}$$
but
$$\sum\limits_{i=1}^{p}i = \frac{p(p+1)}{2}$$
and $\forall p>2$ primes are odd, thus $2 \mid p+1$
which means 
$$\sum\limits_{i=1}^{p}i = \frac{p(p+1)}{2} \equiv 0 \pmod{p}$$
and from $(1)$
$$\sum\limits_{i=1}^{p}i^p \equiv 0 \pmod{p}$$