Prove that all divisors of
$\frac{p^p-1}{p-1}$
are of the form $pk+1$ where $p$ is prime and $k\in\mathbb{Z}$.
Prove that all divisors of
$\frac{p^p-1}{p-1}$
are of the form $pk+1$ where $p$ is prime and $k\in\mathbb{Z}$.
Let $q$ be a prime divisor of $\dfrac{p^p-1}{p-1}$. Then in particular $q$ divides $p^p-1$. But $q\ne p$, and therefore by Fermat's Theorem, $p^{q-1}\equiv 1\pmod{q}$.
So the order of $p$ modulo $q$ divides $q-1$. It also divides $p$. There are two cases to consider: (i) $p$ has order $1$ modulo $q$ and (ii) $p$ has order $p$ modulo $q$.
In Case (i), $p\equiv 1\pmod q$. We have $\frac{p^p-1}{p-1}=p^{p-1}+p^{p-2}+\cdots +1.$ On the right-hand side, there are $p$ terms, each congruent to $1$ modulo $q$. Thus $\dfrac{p^p-1}{p-1}$ is congruent to $p$ modulo $q$, and in particular cannot be divisible by $q$. This contradicts the fact that $q$ is a prime divisor of $\dfrac{p^p-1}{p-1}$.
In Case (ii), the order of $p$ modulo $q$ is $p$. So $p$ divides $q-1$, meaning that $q$ is of the shape $pk+1$.
But any positive divisor of $\dfrac{p^p-1}{p-1}$ is either $1$ or the product of not necessarily distinct primes that divide $\dfrac{p^p-1}{p-1}$, that is, of primes of the shape $pk+1$. And the product of any number of integers of the form $pk+1$ is also of the form $pk+1$. This completes the proof.
Remark: We proved the result for positive divisors of $\dfrac{p^p-1}{p-1}$. In general, it will not be true for negative divisors.