Prime factors greater than $p$ of $N=1+n+n^2+...+n^{p-1}$

Question

Let $p$ be a prime number and $n\ge 1$ be an integer. Prove that the prime factors greater than $p$ of $N=1+n+n^2+...+n^{p-1}$ are of the form $kp+1$, where $k\in \mathbb{N}$.
I observed that $N=\frac{n^p-1}{n-1}$. Now, by Fermat's Little Theorem we know that $n^p \equiv n (\operatorname{mod}p)$, so $n^p-1 \equiv n-1 (\operatorname{mod}p)$. From here we may write $N=\frac{pq_1+r}{pq_2+r}$ from the Euclidean division ($q_1, q_2, r$ are integers). Now I got stuck and I don't know what else to do, the only thing I thought that might work was reducing $N$ modulo $p$ and working in $\mathbb{Z_p}$, but I didn't make much progress.

Answer 1

As you stated, you have

$$N = \frac{n^p - 1}{n - 1} \tag{1}\label{eq1A}$$

Note any prime factors $d \gt p$ of $N$ are not a factor of $n - 1$ since if $n \equiv 1 \pmod d$, then from the summation, you have $p$ terms each congruent to $1$ so they sum up to $p$, giving that $N \equiv p \pmod d$, which is not possible since $p \not\equiv 0 \pmod d$.

This means you have that $d \mid n^p - 1 \implies n^p \equiv 1 \pmod d$. Thus, the multiplicative order of $n$ modulo $d$ is $p$ as the order must divide $p$ and it's not $1$. By Lagrange's theorem, the multiplicative order always divides $\varphi(d)$, i.e., Euler's totient function. For a prime $d$, this would be $\varphi(d) = d - 1$. Thus, as stated in J. W. Tanner's question comment, you have

$$p \mid d - 1 \implies d = kp + 1, \; k \in \mathbb{N} \tag{2}\label{eq2A}$$