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. 
A: Here is a slightly different way to seeing that your fields are contained in cyclotomic ones.
Each root $a_n$ is defined as a root of
$$x^2 - 2 = a_{n-1}$$
The polynomial $x^2-2$ is quite special: it is the degree 2 Chebyshev polynomial, which is to say that it satisfies $f(x+ 1/x) = x^2 + 1/x^2$. Observe that your process can be taken to start at $\alpha_0 = 0 = i + 1/i$. Combining these, we see that $\alpha_1$ can be written as $\zeta+1/\zeta$ where $\zeta^2 = i$; such a $\zeta$ is a root of unity. 
Repeating that process, we see that each $\alpha_n$ can be written as $\zeta + 1/\zeta$ where $\zeta$ is a $2^{n+2}$th root of unity. We know that the sum $\zeta + 1/\zeta$ for a root of unity $\zeta$ can be written in the form $2\cos \theta$, which is exactly what appears in the other answer.
