Let $\mu_n$ be the group of solutions to $x^n - 1$ over $\mathbb{Q}$, and let $\Omega$ be the compositum of all $\mathbb{Q}(\mu_n)$ (in some fixed algebraic closure). I'm trying to understand why $$ \operatorname{Gal}(\Omega/\mathbb{Q}) \simeq \varprojlim (\mathbb{Z}/n\mathbb{Z})^\times $$ I already know that $\operatorname{Gal}(\mathbb{Q}(\mu_n)/\mathbb{Q}) \simeq (\mathbb{Z}/n\mathbb{Z})$, and I know that $\operatorname{Gal}(\Omega/\mathbb{Q})$ is isomorphic to the inverse limit over the Galois groups of all finite Galois subextensions. I'm just having trouble seeing why ALL of the finite Galois subextensions are of the form $\operatorname{Gal}(\mathbb{Q}(\mu_n)/\mathbb{Q}$).
I know that the compositum $\mathbb{Q}(\mu_n) \mathbb{Q}(\mu_k)$ is $\mathbb{Q}(\mu_{\operatorname{lcm}(n,k)})$, and the intersection is just $\mathbb{Q}$ if $\operatorname{gcd}(n,k) = 1$.
How do I see that every finite Galois subextension is cyclotomic?
The comments are correct to point out that this is not true. What is true is that every finite Galois extension in $\Omega$ is in some $\mathbb Q(\mu_n)$
This is one of many examples of how when dealing with infinite sequences of sets, its okay to leave some intermediary sets out as long as all intermediary sets are contained eventually. This is not dissimilar to the following fact about infinite unions:
$$\overline{\mathbb F_p}=\cup_{n\in\mathbb N}\mathbb{F}_p^n=\cup_{n\in\mathbb N}\mathbb{F}_p^{n!}$$
As long as all of the necessary objects get picked out, you don’t need to make sure that the limit is being done in the “finest” way possible. In fact, we often gain a lot by jumping from a very fine listing and a very coarse listing that gives rise to the same object.