Showing Galois Group is Abelian I'm having trouble showing that $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is Abelian. First I want to be able to show that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois, but I'm also not sure how to do this. Any help is appreciated!

It's the splitting field for the polynomial $x^n-1$, hence it is Galois by the characterization of normal extensions as splitting fields for a set of polynomials.
To see it is abelian, it is most direct to show that

$$\text{Gal}\left(\Bbb Q(\zeta_n)/\Bbb Q\right)\cong (\Bbb Z/n\Bbb Z)^\times$$

by showing there is a map of the latter into the former and using the fact that the groups have the same cardinality, since $\varphi(n)=\left|(\Bbb Z/n\Bbb Z)^\times\right|$ by definition and is equal to $[\Bbb Q(\zeta_n):\Bbb Q]$ by direct computation, noting that $\varphi(n)=\text{deg}(\Phi_n(x))$, the $n^{th}$ cyclotomic polynomial, or by showing it is so for $n=p^\alpha$ and using
$$\Bbb Q(\zeta_{p^\alpha})\cap\Bbb Q(\zeta_{q^\beta})=\Bbb Q$$
for distinct primes, $p,q$ and using $[LK:k]=[L:k][K:k]$ when $L\cap K=k$.
The map in the isomorphism is, of course

$$\begin{cases}a\mapsto \sigma_a\\ \sigma_a(\zeta_n)=\zeta_n^a\end{cases}$$

which is easily seen to be a group homomorphism, since
$$\left(\zeta_n^{a}\right)^b=\zeta_n^{ab}$$
and injective since
$$\zeta_n^a=\zeta_n^b\iff n|(b-a)$$
but since $a,b\in(\Bbb Z/n\Bbb Z)^\times$, are represented by integers $1\le a,b\le n-1$ we see that $|b-a|\le n$ with equality only if $b=a$. But the only integer between $-(n-1)$ and $n-1$ which $n$ divides is $0$, since all multiples of $n$ are $nk$ and if $|k|\ne 0$, we have $|nk|> n-1$. So in order for $\zeta_n^a=\zeta_n^b$ it must be that $a=b$, hence injectivity and so we have an isomorphism because the two sets are finite of the same cardinality, so all injections are surjections as well.
