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?

Let $L = \mathbb Q(3^{\frac 1{2^n}})$. 
Why is $L$ algebraic over $\mathbb Q$? It is because the generating set of $L$ is $\mathbb Q \cup \{3^{\frac 1{2^n}}\}$, which are all contained in $\bar{\mathbb  Q}$, the algebraic closure of $\mathbb Q$. Therefore, $L \subset \bar {\mathbb Q}$, which by definition of the algebraic closure implies it is algebraic over $\mathbb Q$.
However, $L$ contains subfields of the form $K_n$, where $K_n = \mathbb Q(3^{\frac 1{2^n}})$. One can conclude that $[K_n : \mathbb Q] = 2^n$, since  $x^{2^n} - 3$ is irreducible via the Eisenstein criterion(shifted).
Then, suppose $L$ were finite, then $[L : \mathbb Q] < \infty $, but by the tower property it is equal to $[L : K_n][K_n : \mathbb Q]$ for each $n$. In particular, $[L : \mathbb Q]$ is a multiple of $2^n$ for all $n$, clearly impossible if it is finite.
