Show adjoining field is not of finite degree

Question

I need to show that the field obtained by adjoining $\sqrt2, \sqrt[\leftroot{-2}\uproot{2}4]{2}, \sqrt[\leftroot{-2}\uproot{2}8]{2}, \sqrt[\leftroot{-2}\uproot{2}8]{2}, . . .,\sqrt[\leftroot{-2}\uproot{2}2^n]{2} $ to $\mathbb{Q}$ is not of finite degree over $\mathbb{Q}$

Answer 1

I assume that you are adjoining $\sqrt[2^n]{2}$ for all $n\geq 1$.

In this case, here is a hint: show that $[\mathbb{Q}(\sqrt[2^n]{2}):\mathbb{Q}]=2^n$ for all $n\geq 1$, and deduce that your field, that I call $L$, will satisfy $[L:\mathbb{Q}]\geq 2^n$ for all $n\geq 1$.

Answer 2

$2^{1/2^n}$ is a unit and has order $2^{n+1}$ in $\Bbb{Z}[2^{1/2^n}]/(3)$.

In particular $\Bbb{Z}[2^{1/2^n}]/(3)$ has at least $2^{n+1}$ elements.

If all the $ \Bbb{Q}(2^{1/2^n})$ are contained in a finite extension $K$ then $\Bbb{Z}[2^{1/2^n}]/(3)$ is a subring of $O_K/(3)$, thus $\Bbb{Z}[2^{1/2^n}]/(3)$ has at most $3^{[K:\Bbb{Q}]}$ elements, which is a contradiction.