For which $p$ primes is $p^{p-1}$ a divisor of $(p-1)^p + 1$?

Question

$p = 2$ and $p = 3$ definitely are solutions. I think these are all the solutions, but how can I prove it?

Answer 1

We consider the given number $N(p) =(p-1)^p+1$ modulo $p^3$. In the following, $p$ is a prime $>3$. Using the binomial formula we have: $$ \begin{aligned} N(p)&\equiv 1+ (-1)^p + p\binom p1(-1)^{p-1} + p^2\binom p2(-1)^{p-2} \\ &= 1+ (-1) + p^2 - p^2\frac 12p(p-1) \\ &= p^2(\text{Number no divisible by $p$})\ . \end{aligned} $$

Answer 2

According to binomial theorem and assuming $p$-odd, thus $(-1)^p=-1$ and $(-1)^{p-1}=1$ $$(p-1)^p+1=\sum\limits_{k=0}^p \binom{p}{k}p^k(-1)^{p-k}+1=\\ -1+\sum\limits_{\color{red}{k=1}}^p \binom{p}{k}p^k(-1)^{p-k} +1=p^2+\sum\limits_{\color{red}{k=2}}^p \binom{p}{k}p^k(-1)^{p-k}=\\ p^2\left(1+\sum\limits_{k=2}^p \binom{p}{k}p^{\color{red}{k-2}}(-1)^{p-k}\right)=...$$ also $\binom{p}{2}=\frac{p(p-1)}{2}$ so we can extract a $p$ from it, thus $$...=p^{\color{red}{2}}\left(1+p\cdot Q\right)$$ and the answer follows.