Q:

When is $\mathbb{Q}(\sqrt{2+\sqrt{2+\sqrt{2 \ldots}}})$ Galois?

Let $\alpha_1=\sqrt{2}$ and inductively define $\alpha_n=\sqrt{2+\alpha_{n-1}}$. For which $n$ is $\mathbb{Q}(\alpha_n)$ Galois? It is easy to see that its Galois closure is generated by the $2^{n}$ combinations of plus and minuses in the nested roots. For $n=3$ these are $\pm \sqrt{2 \pm \sqrt{2 \pm \sqrt{2}}}$. The problem is that all these roots are real, since the limit of $\alpha_n$ is $2$, so there seems to be no easy way to tell if $\mathbb{Q}(\alpha_n)$ Galois.
I have tried to prove that they are always Galois it by induction on $n$, but all I can prove is that (for say $\alpha_n=\sqrt{2+x}$) $\mathbb{Q}(\sqrt{2-x})=\mathbb{Q}(\sqrt{2+x})$, since $\sqrt{2-x}\sqrt{2+x}=\sqrt{4-x^2}$ is a conjugate of $\alpha_{n-1}$ and hence by induction hypothesis it is fixed by $\text{Gal}(L/\mathbb{Q}(\sqrt{2-x}))$, where $L$ is the Galois closure of $\mathbb{Q}(\alpha_n)$. Therefore $\sqrt{2+x}$ is also fixed by $\text{Gal}(L/\mathbb{Q}(\sqrt{2-x}))$.

A:

$$\alpha_n=2\cos\frac{\pi}{2^{n+1}}.$$
Therefore $\alpha_n$ lies in the $2^{n+3}$-th cyclotomic field, which is Galois
over $\Bbb Q$ with Abelian Galois group. Every subextension of
an Abelian extension is Galois. Therefore $\Bbb Q(\alpha_n)/\Bbb Q$ is Galois. 

