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?