While reading a proof of why $\Phi_{n}(X)$, the $n^{th}$ cyclotomic polynomial, is irreducible in $\mathbb{Q}[X]$, I encountered a problem understanding a statement that was made:
If $\zeta$ is a primitive $n^{th}$ root of unity and $p$ a prime that does not divide $n$,
Why is enough to show that $x-\zeta^p$ a linear factor of a polynomial $f(X) \in \mathbb{Z}[X]$ for all such $p$, when you want to prove that $\textrm{ }\forall \textrm{ }a \in (\mathbb{Z}\backslash n\mathbb{Z})^\times$, $x-\zeta^a$ is a linear factor of $f(X)$ ($\zeta$ is a root of $f(X)$)?
Any help will be much appreciated. Thank you!
EDIT
I changed the framing of the question, sorry.
Because you can factorize $a$. Let $f$ be a factor of $\Phi_n$ and let $R$ be the set of roots of $f$. Suppose you knew $\zeta\in R$ implies $\zeta^p \in R$ whenever $p$ is a prime not dividing $n$. Then by induction $\zeta \in R$ implies $\zeta^{p_1p_2p_3\cdots p_r} \in R$ whenever $p_1, \dots, p_r$ are (not necessarily distinct) primes not dividing $n$, which is equivalent to what you want.