Is $F=\Bbb {Q} (\sqrt 2, \sqrt [3]2, \sqrt [4] 2,...)$ different from algebraic closure $\bar {\Bbb {Q}} $ of $\Bbb {Q} $ in $\Bbb {C} $?
Try
I tried to proof by contradiction.
Since $\sqrt 3$ is a root of $x^2-3$, hence $\sqrt 3 \in \bar {\Bbb {Q}} $. Our claim is $\sqrt 3 \notin F$
Assume on the contrary that $\sqrt 3 \in F$. Then $\sqrt 3 \in K=\Bbb {Q} (\sqrt 2, \sqrt [3]2,...,\sqrt [n] 2)\subset F$ for some integer $n$. Now $\sqrt 3=a+b\sqrt [n] 2$ for some $a, b \in \Bbb {Q} (\sqrt 2, \sqrt [3]2,...,\sqrt [n-1] 2)$.
Now $3=a^2+b^2 \sqrt [n]4+2ab\sqrt [n] 2$.
I'm stucked here. Am I on the right track? Please help. Thanks in advance.