Show that $\mathbb{Q}(\sqrt{3},\sqrt[4]{3}, \sqrt[8]{3},...)$ is algebraic over $\mathbb{Q}$ but not a finite extension. 
Show that $\mathbb{Q}(\sqrt{3},\sqrt[4]{3}, \sqrt[8]{3},...)$ is algebraic over $\mathbb{Q}$ but not a finite extension.

I think for the algebraic part, since for every simple extension, each of those elements can be adjoined and each of these simple extensions has a minimal polynomial that cannot be reduced in $\mathbb{Q}$. For example, the simple extension $\mathbb{Q}(\sqrt{3}, \sqrt[4]{3})(\sqrt[8]{3})$ has minimal polynomial $x^{8}-3$. And since each simple extension has an increasingly large degree, the degree of the simple extensions over the previous extension gets larger for each attachment. But I am not sure how to express this formally...
For the infinite degree part, I was thinking because the set $\left \{\sqrt{3},\sqrt[4]{3}, \sqrt[8]{3},...\right \}$ is linearly independent?

Any element $\alpha$ of $F$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $\sqrt[2^k]3$. If in such an expression, $\sqrt[2^n]3$ is one with maximal $k$, then all other $\sqrt[2^k]3$ are powers of $\sqrt[2 k]3$. It follows that $\alpha\in\Bbb Q(\sqrt[2^n]3)$ and $\alpha $ is algebraic.
A different approach for the infinity part: By Eisenstein, the polynomial $X^{2^n}-3$ is irreducible. Hence $[F:\Bbb Q]\ge[\Bbb Q(\sqrt[2^n]3):\Bbb Q]\ge 2^n$, where $n$ is arbitrariy. It follows that $[F:\Bbb Q]$ is inifnite.
