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 $k\in\mathbb{Z}$.
 A: 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.
