Help with proof that if q is a prime divisor of $\frac{n^p-1}{n-1}$, then either q=p or $q\equiv1 \pmod p$ Here is the proof:
 
Why does $(\frac{n^p-1}{n-1},n-1)=(p,n-1)$? Unless n=1 I think it should be the geometric sum $\sum_{i=0}^{p-1} n^i$.
 A: 
Why does $(\frac{n^p-1}{n-1},n-1)=(p,n-1)$?

For any positive integer $k$ we have, $n^k=n^k-1+1=(n-1)(n^{k-1}+n^{k-2}+\cdots +1)+1$, hence, $n^k\equiv 1\pmod{n-1}$. As there are $p$ terms, $\sum_{i=0}^{p-1}n^i\pmod{n-1}=p$ and also note that $gcd(a,b)=gcd(a\pmod{b},b)$. So, we have $$\big(\frac{n^p-1}{n-1},n-1\big)=\Big(\sum_{i=0}^{p-1}n^i,n-1\Big)\\=\Big(\sum_{i=0}^{p-1}n^i\pmod{n-1},~ n-1\Big)\\=(p,n-1)$$
A: Assuming $p$ is and odd prime, this is a result of the more general fact
$$\gcd\left(\frac{a^p + b^p}{a + b}, a + b\right) = \gcd(p, a + b)$$ when $\gcd(a,b) = 1$, which can be shown using the binomial theorem, for instance:
$$\begin{align*}\frac{a^p + b^p}{a + b} & = \frac{((a + b) - b)^p + b^p}{a + b} \\ & = \frac{1}{a + b}\left(b^p + \sum_{i = 0}^p \binom{p}{i}(a + b)^i(-b)^{p - i}\right) \\ & = \frac{1}{a + b}\left(b^p + (-b)^p + \sum_{i = 1}^p \binom{p}{i}(a + b)^i(-b)^{p - i}\right) \\ & = \sum_{i = 1}^p \binom{p}{i} (a + b)^{i - 1}(-b)^{p - i}.\end{align*}$$ All terms in the sum but the first are divisible by $a + b$, so we have $$\frac{a^p + b^p}{a + b} \equiv p(-b)^{p - 1} \equiv pb^{p -1} \equiv p \pmod{a + b},$$ where the last congruence uses the fact $\gcd(b,a + b) = \gcd(b^{p - 1}, a + b) = 1$.
Your special case is $a = n$ and $b = -1$.
