[guided] [step: The minimal polynomial of $\zeta_n$ over $\mathbb{Q}$ is the $n$-th cyclotomic polynomial $\Phi_n(t)$] The $n$-th cyclotomic polynomial is defined by \begin{align*} \Phi_n(t) = \prod_{\substack{1 \le k \le n \\ \gcd(k,n)=1}} (t - \zeta_n^k), \end{align*} so $\Phi_n$ has degree $\varphi(n)$ and $\zeta_n$ is a root. A classical theorem of Gauss shows that $\Phi_n(t) \in \mathbb{Z}[t]$ is irreducible over $\mathbb{Q}$. Because $\Phi_n$ is monic, irreducible in $\mathbb{Q}[t]$, and vanishes at $\zeta_n$, it is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$. Therefore \begin{align*} [\mathbb{Q}(\zeta_n):\mathbb{Q}] = \deg \Phi_n = \varphi(n). \end{align*} [step: $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois] The field $\mathbb{Q}(\zeta_n)$ is the splitting field of $t^n - 1$ over $\mathbb{Q}$: every root of $t^n - 1$ is a power $\zeta_n^k$ for some $0 \le k \le n-1$, so all roots already lie in $\mathbb{Q}(\zeta_n)$. Moreover $t^n - 1$ has $n$ distinct roots in $\mathbb{C}$ (its derivative $nt^{n-1}$ shares no common root with it), so the extension is separable. A splitting field of a separable polynomial over $\mathbb{Q}$ is a Galois extension, hence $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois. $\blacksquare$ [/guided]