If ($b^n-1$)/($b-1$) is prime, prove $n$ is prime relatively prime to $b-1$ Suppose we have a prime of the form $\left(\cfrac{b^n-1}{b-1}\right)$ with ${b, n} > 1$. Show that: 
$n$ is prime.
$n$ is relatively prime to $b-1$.
The first part is easily proven, but I cannot come of with a proof for the second part. Anyone else?
 A: Using the fact: $\cfrac{x^n-1}{x-1} = \sum_{i= 0}^{n-1} x^i$
Proof by contradiction:
If $n$ is not relatively prime to $b-1$ and since $n$ is a prime, then $b-1 \equiv 0 \pmod{n} \implies b \equiv 1 \pmod {n}$
$\cfrac{b^n-1}{b-1} \equiv  \sum_{i= 0}^{n-1} b^i \equiv  \sum_{i= 0}^{n-1} 1^i \equiv n\equiv 0 \pmod{n}$
So the $\left(\cfrac{b^n-1}{b-1}\right)$ is not a prime which contradict to the condition.
A: An alternative proof, which I think is not nearly as elegant as Zack's:
Suppose that $b^n -1 = p(b-1)$, where $p$ is prime.
Then, rewrite $b^n = ((b-1)+1)^n$. Expanding with the binomial theorem, we see that the first formula can be rewritten as
$$(b-1)^{n-1} + n S(b) =p$$
Where $S(b)$ is a summation which is easily found through using the binomial theorem, but I will spare you the details. But now we are done: if either of them shares a common divisor, then we would see that $p$ would have to be a composite number because we could factor their divisor out of the sum. Thus, $b-1$ and $n$ are relatively prime.
