Remainder of Fermat's little theorem sum

Question

With p being a prime number, what is the remainder of $$\sum_{k=1}^{p-1} {k^{p-1}}$$ divided by p ?

I know that Fermat's little theorem states that for a prime number p, and a number A that is not divisible by p: $${A^{p-1}}≡ 1\mod p $$

So I'm thinking that the remainder would be 1, what am I missing here?

Any guidance, clues or solutions are much appreciated, thanks!

Answer 1

So I'm thinking that the remainder would be 1, what am I missing here?

What you are missing, is the basic rule of $$\underset{p-1 \text{ times}}{\underbrace{1+1+\cdots+1+1}}=p-1$$ still working in modular arithmetic. It simply wraps if it gets too big, which this sum isn't.

Answer 2

The logic seems fine, but I don't think your conclusion follows from it. From what you've shown, $$\sum_{k=1}^{p-1}k^{p-1}\equiv\sum_{k=1}^{p-1}1 = p-1 \equiv -1\pmod p$$

I'm pretty sure you don't get $1$ from simplifying that.