Prove congruence using Fermat's Little Theorem I'm trying to prove that if $(p - 1)\mathrel{\mid} n$ where $p$ is prime and $n ≥ 1$, then
$$1^n + 2^n + \dotsb + (p - 1)^n + p^n ≡ -1 \pmod p.$$
My attempt thus far:
If $(p - 1)\mathrel{\mid} n$ then $n = (p - 1)d$ for some $d ≥ 0$, so we can substitute and obtain
$$1^n + 2^n + \dotsb + (p - 1)^n + p^n ≡ 1^{(p - 1)d} + 2^{(p - 1)d} + \dotsb + (p - 1)^{(p - 1)d} + p^{(p - 1)d}.$$
Note $p$ cannot divide $p - 1$, nor can $p$ divide any value less than $p$ (i.e. cannot divide $1, 2, \dotsc, p - 2$). Using Fermat's little theorem and simplifying (mod $p$) this congruence becomes
$$1^d + 2^d + ... + (p - 1)^d + 0.$$
And from here I'm lost. I'm not even sure if I'm on the right track here. I'd appreciate any help or suggestions.

The problem is that  you simplified in a wrong way. You have $j^{p-1}$ is $1$ modulo $p$. Not $j$ as you wrote. Thus you get $1^d + 1^d + \dots + 1^d + 0$. 
