It is well known that in characteristic p we have the "freshman dream" $x^p -1 = (x - 1)^p$. Some experimentation seems to suggest that the heuristic computation $(x-1)(x-1)^{p-1} = (x-1)^p = x^p - 1 = (x-1)(1 + x + \dots + x^{p-1}) \implies (x-1)^{p-1} = 1 + x + \dots + x^{p-1}$ actually yields a correct result.
Playing around with binomial coefficients leads me to believe that $ p-1 \choose k$ = $(-1)^k$ mod p, which would imply the result, but I do not see how to show this. It seems like there should be an easy way, I just do not see it.