Using Fermat's Little Theorem, prove $1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv -1 \pmod p$ where $p$ is a prime.
So we would use Fermat's theorem: $a^{p-1}\equiv 1\pmod p$
Would the proof go something like this....?
$1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv 1+\ldots+1$
There would be $p-1$ many $1$'s so (i.e. $p-1=5-1=4$ so $1+1+1+1=4$)
$p-1 \equiv -1\pmod p$
$p \equiv 0 \pmod p$
Would that be right?