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Let $ (a_n)_{n \in \mathbb{Z\geq0}} $ with $a_0 = 2 $ and $a_{n+1}= \sqrt{a_n}$ with $a_{n+1}>0$. We need to show that $[\mathbb{Q}(a_n):\mathbb{Q}]=2^n $$\forall n \in \mathbb{Q}$.

To prove this we need to show that $\mathbb{Q}(a_n) \subset \mathbb{Q}(a_{n+1})$ and $[\mathbb{Q}(a_{n+1}):\mathbb{Q}(a_n)]=2$ trough induction.

Now the easiest part is to show that $\mathbb{Q}(a_n) \subseteq \mathbb{Q}(a_{n+1})$ and I managed to do this. The tough part is to show that $\mathbb{Q}(a_n) \subset \mathbb{Q}(a_{n+1})$. If anyone can help me out here I would very much appreciate it.

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  • $\begingroup$ You can draw all conclusions you wish from the explicit solution of the recursion, which is $a_{n}=(a_{0})^\frac{1}{2^n}$ $\endgroup$ – Dr. Wolfgang Hintze Mar 12 at 14:17
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I think it's easier than you think. $a_n$ is a root of $$ x^{2^n} - 2 $$ which, by Eisenstein's criterion, is irreducible over the rational numbers.

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