I came across the proof of the following fact in the Chapter $2$, Lemma $2.9$ of Lawrence CWashington's book, Introduction to Cyclotomic Fields.
Let $p$ be a rational prime, $n\in\mathbb{N}$ and $a\in\mathbb{Z}$. Let $p\nmid a$ and $p\nmid n$. Then, $p \mid \Phi_{n}(a)$ if and only if the order of $a$ modulo $p$ is $n$. Here, $\Phi_{n}$ is the $n^{th}$ cyclotomic polynomial.
I think the proof could be generalized to a slightly general statement: Let $\alpha\in\mathbb{Z}[\zeta_{n}]$ and $\mathfrak{p}$ be a prime in $\mathbb{Z}[\zeta_{n}]$ that neither divides $\alpha$ nor $n$ in $\mathbb{Z}[\zeta_{n}]$. Then, $\mathfrak{p}\mid\Phi_{n}(\alpha)$ if and only if the order of $\alpha$ modulo $\mathfrak{p}$ is $n$.
I think that a mere copying of that proof works for this case too. however, I will reproduce the proof so that it would be easier to respond. Is this really correct? I have tried my best and it does appear correct.
Proof Since $x^{n} - 1 = \prod_{d\mid n} \Phi_{d}(x) $, $\mathfrak{p}\mid\Phi_{n}(\alpha)$ gives that $\alpha^{n} \equiv 1 (\text{mod } \mathfrak{p})$. If the order of $\alpha$ modulo $\mathfrak{p}$ was $k\mid n$ with $k < n$, then using $$ x^{n}-1 = \Phi_{n}(x) \Phi_{k}(x) (\text{ some other factors }) $$ we have that $\alpha^{n} \equiv 1 (\text{mod } \mathfrak{p}^{2})$.
Using the fact that for every polynomial $f(x)$ and every $r \in\mathfrak{p}$ we have $f(x+r) \equiv f(x) (\text{mod } \mathfrak{p})$ (and using it for both $\Phi_{n}$ and $\Phi_{k}$), we must have that for every $r\in\mathfrak{p}$,
$$(\alpha + r)^{n} \equiv 1 (\text{mod } \mathfrak{p}^{2})$$ giving after simplification that
$$rn\alpha^{n-1} \equiv 0 (\text{mod } \mathfrak{p}^{2}),$$ which is a contradiction because $\mathfrak{p}$ neither divides $\alpha$ nor $n$ in $\mathbb{Z}[\zeta_{n}]$.
The other direction is easier. If the order of $\alpha \text{mod } \mathfrak{p}$ is $n$, then $\mathfrak{p}\mid(\alpha^{n}-1)$ which gives that $\Phi_{d}(\alpha) = 0$. If $d \neq n$ then we would end up having that $$\alpha^{d} - 1 = \prod_{r \mid d} \Phi_{r}(\alpha)$$, which would contradict that order of $\alpha (\text{mod } \mathfrak{p})$ is $n > d$.