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Prove, if $p$ is an odd prime: $1^{p-1} + 2^{p-1} + ... + (p-1)^{p-1}$ is congruent to -1 modulo p.

I was going to try and show this by induction but I am having trouble with the base case that $1^{p-1}\equiv-1$ mod p. This makes me think that induction isn't the right way to go. Any help would be much appreciated. Thank you!

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    Fermat's Little theorem ? ... – 2017-02-20

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Use Fermat's little theorem to prove that each term is congruent with $1$ modulo $p$. Then the whole sum is going to be congruent with $p-1=-1$ modulo $p$

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    Doesn't the theorem say that $a^{p-1}\equiv a$ mod p though?... – 2017-02-20
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    @Math4Life It says $a^p\equiv a\pmod{p}$ dividing through $a$ you get $a^{p-1}\equiv 1\pmod{p}$ – 2017-02-20
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    Thank you. Since I will have 1+1+...+1 p-1 times it will give -1. I see it now I think. Thanks! – 2017-02-20