Some lemma ahead:
Lemma Let $E/F$ be the field extension. Then all F-automorphisms defines a permutation of roots of any polynomials in $F$
Proof For any root $\alpha$ of any polynomial $p(x) \in F[x]$, it is straightforward that $$ p(\sigma(\alpha)) = \sigma(p(\alpha)) = 0$$ for any $\sigma \in \text{Gal}(E/F)$. Therefore $\sigma(\alpha)$ is indeed a root of $p(x)$.
The proof of irreducibility of $\Psi_n(x)$ over $\mathbb{Q}[x]$: $$\Psi_n(x) := \prod_{(i,n)=1}(x - \zeta^i_n)$$ It is proven $\Psi_n(x) \in \mathbb{Z}[x]$. Consider the group of nth roots of unity, cyclic and isomorphic to $\mathbb{Z}_n$.
First note that any $\mathbb{Q}$-automorphism must permute the group of nth roots of unity (actually defines an automorphism of that group). It is easy to check any automorphism on $\mathbb{Z}_n$ can and can only map a unit to another. Likewise any $\mathbb{Q}$-automorphism must map a primitive to another primitive, and we can identify any $\mathbb{Q}$-automorphism: $\zeta_n \mapsto \zeta_n^r, (n, r) = 1$ solely by r.
By the lemma, either none of the elements of the primitive nth roots of unity is a root of a polynomial, or all of them. In other words, a minimum polynomial for any $\zeta^i_n$ must have degree more than $\Psi_n$, therefore must be $\Psi_n$. By the definition of minimum polynomial, $\Psi_n$ is irreducible.
Is this proof correct?