= Solution
Fix a primitive $n$th root of unity $\zeta_n$. The $n$th <cyclotomic polynomial> is
$$
\Phi_n(X)=
\prod_{\substack{1\leq a\leq n\\(a,n)=1}}
(X-\zeta_n^a).
$$
Its roots are exactly the roots of unity of order $n$.
Partitioning all $n$th roots of unity by exact order gives the <cyclotomic factorization>
$$
X^n-1=\prod_{d\mid n}\Phi_d(X).
$$
We prove $\Phi_n\in\mathbb Z[X]$ by induction on $n$. The assertion is clear for $n=1$. If it holds for every proper divisor of $n$, then
$$
\Phi_n(X)=\frac{X^n-1}{\prod_{d\mid n,\ d<n}\Phi_d(X)}.
$$
The denominator is monic and belongs to $\mathbb Z[X]$. Exact polynomial division shows that the quotient lies in $\mathbb Q[X]$, and Gauss's lemma makes this monic quotient an element of $\mathbb Z[X]$.
Now let $K\subseteq\mathbb C$. The minimal polynomial of $\zeta_n$ over $K$ divides $\Phi_n$, so every $K$-conjugate of $\zeta_n$ is another primitive root $\zeta_n^a$ and lies in $K(\zeta_n)$. The extension is therefore normal. It is separable because the characteristic is zero, and hence it is Galois.
Every $\sigma\in\operatorname{Gal}(K(\zeta_n)/K)$ has
$$
\sigma(\zeta_n)=\zeta_n^a
$$
for some $a\in(\mathbb Z/n\mathbb Z)^\times$. Since $\zeta_n$ generates the extension, this gives an injective homomorphism
$$
\operatorname{Gal}(K(\zeta_n)/K)
\hookrightarrow(\mathbb Z/n\mathbb Z)^\times.
$$
The group on the right is abelian, so the Galois group is abelian, as summarized by the <cyclotomic field> construction.
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