Let $E = \mathbb{Q}\big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$. Prove $[E: \mathbb{Q}]$ is not finite Problem: Consider the subfield $$E = \mathbb{Q} \big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$$ of $\mathbb{R}$. Prove that $[E: \mathbb{Q}]$ is not finite.
Attempt: I proved that $E$ is algebraic over $\mathbb{Q}$. Because let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$. Then $f(x) = x^n - 2 \in \mathbb{Q}[x]$ is the minimal polynomial of $u$ over $\mathbb{Q}$ such that $f(u) = 0$. Hence $E$ is algebraic over $\mathbb{Q}$.
I now want to prove the degree of extension is not finite. I assume it is finite. Suppose then $[E: \mathbb{Q}] = k \in \mathbb{N}$. I don't know how to derive a contradiction from this. 
I know that if $u$ is algebraic over $\mathbb{Q}$ and $f$ is the minimal polynomial of $u$ over $\mathbb{Q}$, then $[\mathbb{Q}(u) : \mathbb{Q}] = \deg(f)$. 
Help/suggestions are appreciated.

Note that the minimal polynomials of all the different roots of $2$ have ever increasing degrees. Therefore a finite extension cannot contain all of them. Specifically, if $E$ is a finite extension with $[E:\Bbb Q]=r$, then $\sqrt[r+1]2\notin E$.
