Prove that the splitting field of $X^n - 1$ has degree $\phi(n)$ over $\mathbb{Q}$, where $\phi(n)$ denotes the number of integeres which are less than n and are co-prime to n.
I understand that the splitting field is $\mathbb{Q}(\omega)$, where $\omega$ is the primitive $n^{th}$ root of unity.
I am not able to proceed further. In case $n$ is a prime it is clear from the answer below
Prove that the degree of the splitting field of $x^p-1$ is $p-1$ if $p$ is prime
This follows because the irreducible polynomial of $n$th roots of unity is $$ \prod_{d\mid n}(X^d-1)^{\mu(n/d)}$$ (e.g., by the principle if inclusion/exclusion: $X^n-1$ has all $n$th roots, for $p\mid n$, you divide out $X^{n/p}-1$ to remove roots that are already $n/d$th roots; but that removes $n/(p_1p_2)$th roots twice, hence you multiply back with $X^{n/(p_1p_2)}-1$, etc.).
Hence its degree is $$ \sum_{d\mid n}\mu(n/d)d=\phi(n).$$