Consider $$K:=\Bbb{Q}(\{\root n\of2\mid n≥2\})⊆\Bbb{R},$$ find $Gal(K/\Bbb{Q})$.
So it would be at most $n!$ (infinite) number of $\Bbb{Q}$- automorphism over K. I should apply the simple extension theorem adding one by one element of $K$ but I know some of the roots get out of the precedesor field. So maybe I could use induction or maybe I can simplify K.
Thats exactly what happens with $Gal[\Bbb{Q}(2^{1/2},2^{1/4})/\Bbb{Q}]$ that is basically the same that $Gal[\Bbb{Q}(2^{1/4})/\Bbb{Q}]$ and this is basically because $\Bbb{Q}(2^{1/2},2^{1/4})= \Bbb{Q}(2^{1/4})$ so I wish I can simplify $K$ like this too and I think the number of Q-automorphisms may be finite.
Thanks.
It is still not clear to me what the field $K$ is supposed to be. But assuming that $K$ is the field gotten by adjoining all the real zeros of all the polynomials $x^n-2, n\ge2$, to $\Bbb{Q}$, then the conlusion is that
This is seen as follows. Let $\sigma$ be an automorphism of $K$. I claim that $\sigma(2^{1/n})=2^{1/n}$ for all $n\ge2$. Let's fix $n\ge2$. The polynomial $x^{2n}-2$ has two zeros in $K$, namely $\pm z_n$ with $z_n=2^{1/2n}$. We can thus deduce that $\sigma(z_n)=\pm z_n$. Therefore $\sigma(z_n^2)=z_n^2$, which is exactly the claim that $$\sigma(\root n\of2)=\root n\of2.$$ Because $K=\Bbb{Q}(\{\root n\of2\mid n\ge2\}$), the claim follows.