Some proofs for Cyclotomic extension $\Bbb K(\xi _n)/ \Bbb K $ I need to proof the following:
 $\Bbb K$ is  a field with $Char(\Bbb K)=0$ or $Char(\Bbb K )=p$ (when $p$ is a prime).
(1) for $n \in \Bbb K^\times$ there is always a primitive unit root in the algebraic closure:   $\xi_n \in \overline {\Bbb  K}$.
(2)The extension $\Bbb K (\xi_n)/\Bbb K $ is a Galois extension.
(3) there is embedding $(Gal(\Bbb K (\xi_n)/\Bbb K) \hookrightarrow (\Bbb Z/n\Bbb Z)^\times  $
 A: Existence of primitive root of unity
When $p \nmid n$, the polynomial $X^n-1$ is separable.
Therefore, there are $n$ distinct $n$-roots of unity in $\overline K$.
Now we prove inductively that for any $d$ dividing $n$, there are $\varphi(d)$ primitive $d$-roots of unity, bearing in mind the formula $\sum_{i \mid d} \varphi(i) = d$, where $\varphi$ is Euler's totient function.
In the induction step: noting that $X^n-1 = (X^d-1)(X^{n-d} + \cdots + X^{2d} + X^d + 1)$, we see that there are $d$ $d$-roots of unity. Therefore:
$$\begin{array}{rcl}
d
&=& \#\text{$d$-roots of unity} \\
&=& \sum_{i \mid d} \#\text{primitive $i$-roots of unity} \\
&=& \#\text{primitive $d$-roots of unity} + \sum_{i \mid d, i<d} \varphi(i) \\
&=& \#\text{primitive $d$-roots of unity} + \sum_{i \mid d} \varphi(i) - \varphi(d) \\
&=& \#\text{primitive $d$-roots of unity} + d - \varphi(d) \\
\end{array}$$
thus estbalishing the result.
And then the result follows from the fact that $\varphi(n) > 0$.

$K(\zeta_n)/K$ is Galois
It is the splitting field of $X^n-1$, so it is normal.
Separability is already discussed.

$Gal(K(\zeta_n)/K) \hookrightarrow (\Bbb Z/n\Bbb Z)^\times$
Each element of $Gal(K(\zeta_n)/K)$ sends $\zeta_n$ to $\zeta_n^m$ for some $m$, which uniquely determines the element. The first fact is because the element must send a root of $X^n-1$ to another root of the same polynomial. This gives an injection into $\Bbb Z/n\Bbb Z$.
To see that the image is contained in the units, note that if the inverse of the element sends $\zeta_n$ to $\zeta_n^j$, then $mj = 1 \pmod n$.
For a more categorical approach in the same vein:
$$Gal(K(\zeta_n)/K) \hookrightarrow \operatorname{Aut}(C_n) \equiv (\Bbb Z/n\Bbb Z)^\times$$
