The Question is below:
For which primes $p$ is $(p − 1)^p + 1$ a power of $p$?
I think the answer is $2$ and $3$, none of the others work.
Here is what I have:
let $p^k-1=(p-1)^p$. Then we have $$p^{k-1}+...+p+1 \equiv0 \pmod{p-1}.$$
Please also include a proof of your answer
On
Suppose $(p-1)^p=p^n-1$ for some prime $p$ and some positive integer $n$.
Observe that $(p,n)=(2,1),(3,2)$ are solutions. We claim there aren't any more.
Suppose $p \ge 5$. From the Binomial Theorem,
$$ (p-1)^p = p^n-1 = \big((p-1)+1\big)^n - 1 = \sum_{k=1}^n {n \choose k} (p-1)^k, $$
so that
$$ (p-1)^{p-1} = \sum_{k=0}^{n-1} {n \choose k} (p-1)^k. $$
But now the LHS is a multiple of $p-1$ whereas the RHS is $1$ more than a multiple of $p-1$. Since $p-1 \ne 1$, this is impossible.
Therefore,
$$ (p-1)^p+1 = p^n \Longleftrightarrow (p,n) = (2,1)\:\text{or}\:(3,2). $$
If $(p-1)^p+1=p^k$ then
\begin{align*} k&=\log_p \left(1+(p-1)^p\right)\\ &=\log_p \left(p^p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)\right)\\ &=p+\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right) \end{align*}
clearly
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)<0$$
since $\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p<1$
and so it suffices to show that
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)>-1$$
for $p>3$ since then $p+\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)$ cannot be an integer for $p>3$. Differentiating $\left(1-\frac{1}{x}\right)^x$ we get that it is an increasing function and so for $p\geq5$ we get that
$$\left(1-\frac{1}{x}\right)^x\geq\left(1-\frac{1}{5}\right)^5>.3$$
for $p\geq5$ we also clearly have that $\frac{1}{p}\leq \frac{1}{5}=.2$: it is thus that
\begin{align*} \frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p &> \left(1-\frac{1}{p}\right)^p\\ &>.3\\ &>.2\\ &\geq \frac{1}{p} \end{align*}
taking logs base $p$ of both sides gets that
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)>-1$$
and our proof is complete.