If $p$ and $q$ are primes such that $q \mid {\frac{x^p-1}{x-1}}$ then prove that $q\equiv 1 \pmod{p}$ or $q\equiv 0 \pmod{p}$. 
QUESTION: If $p$ and $q$ are primes such that $$q \mid {\frac{x^p-1}{x-1}} ,  (x\in\Bbb{N}, x>1)$$ then prove that $q\equiv 1 \pmod{p}$ or $q\equiv 0 \pmod{p}$.


MY ANSWER: I came across this lemma, but couldn't prove the second part properly. Here's what I did -
By Fermat's Little Theorem we know that $x^p\equiv{x}\pmod{p}$. Therefore, $$\frac{x^7-1}{x-1}\equiv\frac{x-1}{x-1}=1\pmod{7}$$ Therefore, $q\equiv{1}\pmod{7}$.
Now, I cannot prove that $q\equiv{0}\pmod{7}$. Not simultaneously ofcourse, I know that ..
Here's my try -
We can write the above equation as $$x^6+x^5+x^4+x^3+x^2+x+1$$ But what after this? Even if I chose $q=7$, it does not divide the above equation for all values of $x$. Say $x=7$, then the equation can be rewritten as $$7^6+7^5+7^4+7^3+7^2+7+1$$ and $$7\nmid{7^6+7^5+7^4+7^3+7^2+7+1}$$ So, how do I rigorously prove that $q\equiv{0}\pmod{p}$ ? Or, for which cases is this true?
Any help will be much appreciated. Thank you :)

EDIT: Terrible Mistake :P The first proof of $q\equiv{1}\pmod{7}$ is wrong. So, now I am left with a full question to be proved °_°

 A: There are $2$ cases to consider.

Case $1$: $x \equiv 1 \pmod{q}$
Dividing $x - 1$ into $x^p - 1$, and using $x \equiv 1 \pmod{q}$, gives
$$0 \equiv \sum_{i=0}^{p-1}x^{i} \equiv \sum_{i=0}^{p-1}1^{i} = p \pmod{q} \tag{1}\label{eq1A}$$
Since $p$ and $q$ are primes, this means $p = q$, i.e., $q \equiv 0 \pmod{p}$.

Case $2$: $x \not\equiv 1 \pmod{q}$
In this case, you have
$$x^p - 1 \equiv 0 \pmod{q} \implies x^p \equiv 1 \pmod{q} \tag{2}\label{eq2A}$$
The multiplicative order of $x$ modulo $q$ divides any power of $x$ which is congruent to $1$. Since $x \not\equiv 1 \pmod{q}$, this means the multiplicative order must be $\gt 1$. As $p$ is prime, its only factors are $1$ and $p$, so this means the multiplicative order of $x$ modulo $q$ must be $p$.
However, Fermat's little theorem states
$$x^{q-1} \equiv 1 \pmod{q} \tag{3}\label{eq3A}$$
This means $p \mid q - 1$, i.e.,
$$q \equiv 1 \pmod{p} \tag{4}\label{eq4A}$$

In summary, this shows
$$q \equiv 1 \pmod{p} \; \; \text{ or }  \; \; q \equiv 0 \pmod{p}  \tag{5}\label{eq5A}$$
A: Generalizing this answer first suppose that $q \not \equiv 1\pmod{p}$ so that $\gcd(q-1, p) =1$. Then there exists integers $n,m$ such that $pn + (q-1)m =1$ by Bezout's lemma. Then we have $$x \equiv x^{pn +(q-1)m} \equiv (x^p)^n(x^{q-1})^m \equiv 1 \pmod{q} $$
by Fermat's little theorem and the fact that $q|x^p-1.$
Hence $$\frac{x^p-1}{x-1} = 1+x+\dots+x^{p-1} \equiv p \pmod{q}.$$ This shows that $q=p$. The contrapositive says $q \ne p$ implies $q \equiv 1 \pmod{p}$, so we're done since either $q=p$ or $q \ne p$.
