In a Wikipedia article on roots of unity modulo n, in the section on roots of unity, it is stated:
If $x$ is a $k$-th root of unity and $x-1$ is not a zero divisor, then $\sum_{j=0}^{k-1} x^j \equiv 0 \pmod{n}$, because $(x-1) \cdot \sum_{j=0}^{k-1} x^j \equiv x^k -1 \equiv 0 \pmod{n}$.
When I first read it, I accepted it, the equation makes sense and is plausible.
But, on re-reading, I found that $x-1$ is not a zero divisor to be a strong condition, if we assume that we are "only" dealing with roots of unity. Because assuming we know that $x^k \equiv 1 \pmod{n}$ is a $k$-th root of unity, how should it follow that $x-1$ is not a zero divisor?
If, on the other hand, primitive roots were assumed, then I could accept that $(x-1)$ is not a zero divisor, because here it is true that firstly $x^k \equiv 1$ and secondly $x^j \not\equiv 1$ for $j=1,...,k-1$.
In a nutshell: I am interested (if possible) in how to deduce from $x^k \equiv 1 \pmod{n}$ that $x-1$ is not a zero divisor.
I hope my question has the appropriate motivation to be discussed here. I look forward to your answers and thank you in advance!