Is $every$ prime factor of $\frac{n^{163}-1}{n-1}$ either $163$ or $1\;\text{mod}\;163$? This was inspired by this question. More generally, given prime $p$ and any integer $n>1$, define,
$$F(n) = \frac{n^p-1}{n-1}=n^{p-1}+n^{p-2}+\dots+1$$

Q: Is every prime factor of $F(n)$ either $p$, or $1\;\text{mod}\;p$?

 A: Yes: suppose $q$ divides $(n^p-1)/(n-1)$. Then $q$ divides $n^p-1$ so $n$ has order either $1$ or $p$ modulo $q$. In the second case $p$ divides $q-1$, so $q \equiv 1$ mod $p$. Otherwise $n = 1 + \lambda q$ for some $\lambda \in \mathbb{N}$ and we have
$$ \frac{n^p-1}{n-1} = \frac{(1+\lambda q)^p-1}{\lambda q} = \sum_{r=0}^{p-1} \lambda^r q^r \binom{p}{r+1}. $$
The first summand is $p$ and all the rest are divisible by $q$. Since $q$ divides the left-hand side, we have $q=p$.
Generalization. $(n^p-1)/(n-1) = \Phi_{p}(n)$, where $\Phi_d(x) \in \mathbb{Z}[x]$ is the $d$th cyclotomic polynomial. In a comment on the question Wojowu mentioned the following generalization: if $d \ge 2$ and $q$ is a prime dividing $\Phi_d(n)$, then either $q$ divides $d$, or $q \equiv 1$ mod $d$. Since it's a nice result, and could be more widely known, I've added a proof below. In particular, it can be used to show that any arithmetic progression $1,1+d,1+2d,1+3d,\ldots $ contains infinitely many primes.
Proof. Since $\Phi_d(x) \mid x^d-1$, we have $q \mid n^d-1$. Take $s$ maximal such that $q^s$ divides $n^d-1$. Suppose that $n$ has order $t$ modulo $q^s$. If $t=d$ then, since the group of units of $\mathbb{Z}/q^s\mathbb{Z}$ has order $\phi(q^s) = q^{s-1}(q-1)$, it follows from Lagrange's Theorem that $d \mid q^{s-1}(q-1)$. Hence either $d \mid q-1$ (and so $q \equiv 1$ mod $d$), or $d$ has a common factor with $q^{s-1}$, so $q \mid d$.
We now rule out the remaining case, where $t$, the order of $n$ modulo $q^s$, properly divides $d$. Since the polynomials $x^t-1$ and $\Phi_d(x)$ are coprime, and $x^t-1 \mid x^d-1$, we have
$$x^d-1 = (x^t-1)\Phi_d(x)f(x) $$
for some $f(x) \in \mathbb{Z}[x]$. Substituting $n$ for $x$ we get $n^d-1 = (n^t-1)\Phi_d(n)f(n)$. Since $q^s \mid n^t-1$ and $q \mid \Phi_d(n)$, it follows that $q^{s+1} \mid x^d-1$, contrary to the choice of $s$. $\Box$
A: Community Wiki answer, expanding that by Mark Wildon.
We have a fixed prime $p=163,$ but we will keep the symbol $p.$ The main tool here is Fermat's Little Theorem, which says that, for a prime $q$ that does not divide $n,$ we get
$$ n^{q-1} \equiv 1 \pmod q. $$
Furthermore, there is some minimal positive integer $t,$ sometimes called the "order," such that $n^t \equiv 1 \pmod q.$ Manipulations using the gcd show that this $t |(q-1).$
We have some $q$ dividing $n^p - 1.$ So $n^p \equiv 1 \pmod q.$ We have some order $t.$ This says that $t | p.$ Either $t=1$ or $t=p.$
If the order $t=p,$ this means $p | (q-1).$ This is main conclusion asked about.
If the order $t=1,$ this means $n \equiv 1 \pmod q.$ We begin by expanding $n=1 + \lambda q.$ But then Mark gives
$$ \frac{n^p-1}{n-1} = \frac{(1 + \lambda q)^p - 1}{\lambda q}  $$ This is further expanded as
$$ (1/\lambda q) (1 + p \lambda q + \binom{p}{2} \lambda^2 q^2 + \cdots + p  \lambda^{p-1} q^{p-1} + \lambda^p q^p - 1  ) $$
or
$$ p + \binom{p}{2} \lambda q + \binom{p}{3} \lambda^2 q^2 + \cdots + p  \lambda^{p-2} q^{p-2} +  \lambda^{p-1} q^{p-1}. $$
All but the first term of this have an explicit $q$ factor. The missing item is the first term, $p.$ If $q$ divides the whole thing, then $q$ divides $p$ and $q=p.$
