Neukirch makes the following assertion in Algebraic Number Theory:
Let $L = \mathbb{Q}(\zeta)$ where $\zeta$ is a primitve $n$th root of unity. Let $p$ be an integer coprime to $n$. For any prime ideal $\frak{p}$ lying over $p$ in $L$ we see that ...
"passing to the quotient $O_L \rightarrow \frac{O_L}{\frak{p}}$ maps the group $\mu_n$ of the $n$th roots of unity bijectively onto the group of $n$th roots of unity of $\frac{O_L}{\frak{p}}$."
Unfortunately he is sparse on the details and I fail to even see why the $n$th roots of unity in the latter field are even of the same cardinality as the $n$th roots of unity in $O_L$. Is there some easy way to see this bijection? Before this he mentions that the polynomial $X^n - 1 \mod \frak{p}$ has no multiple roots, so maybe that has something to do with it?
Note that $ \prod_{k=1}^{n-1} (1- \zeta^k) = n $. (Neukirch uses a similar identity to get $p$ as a product of $1- \zeta^k$ with only primitive roots of unity)
Since $p$ coprime to $n$, and $P \vert p$ this means $ \prod_{k=1}^{n-1} (1- \zeta^k) \not\equiv 0 (\textrm{mod}\ P)$ , and thus $\zeta^k \not\equiv 1$ for any $k$. Therefore the map $\zeta^k \rightarrow (\zeta^k \mod P)$ is injective. Noting there cannot be more than $n$ roots of $X^{n} - 1 \mod P$, we have all of them.