cardinality of Galois group in $\mathbb{Q}$($\zeta_n$) Let $\mathbb{Q}$($\zeta_n$) be some cyclotomic field, where $\zeta_n$ is a n-th root of unity.

I already managed to show that $\mathbb{Q}$($\zeta_n$) is an Galois extension, but now i struggle to show that 
$[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$, where $\varphi(n)$ is Euler's totient function.
For any $\sigma \in Gal(\mathbb Q(\zeta_n)/\mathbb Q)),$ $\sigma(\zeta_n)$ must be a primitive $n$-th root of unity, because $\sigma$ is a $\mathbb Q$-automorphism. Since $\zeta_n$ generates $\mathbb Q(\zeta_n)/\mathbb Q$, it follows that this group has as many elements as the set of primitive $n$-th roots of unity, that is $\varphi(n)$, where $\varphi$ denotes Euler's totient function.
For any $\sigma \in Gal(\mathbb Q(\zeta_n)/\mathbb Q)),$ $\sigma(\zeta_n)$ must be a primitive $n$-th root of unity, because $\sigma$ is a $\mathbb Q$-automorphism. Since $\zeta_n$ generates $\mathbb Q(\zeta_n)/\mathbb Q$, it follows that this group has as many elements as the set of primitive $n$-th roots of unity, that is $\varphi(n)$, where $\varphi$ denotes Euler's totient function.