Question: **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.

Answer: 
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 $\sqrt[r+1]2\notin E$.
Your starting point is wrong: if $u\in E$, there's no reason why $u^n=2$, for some $n$.

However, there is $m$ such that
$$
u\in \mathbb{Q}(\sqrt{2},\sqrt[3]{2},\dots,\sqrt[m]{2})
$$
because
$$
E=\bigcup\_{n>1}\mathbb{Q}(\sqrt{2},\dots,\sqrt[n]{2})
$$
(prove it).

Now, $\sqrt[k]{2}\in\mathbb{Q}(\sqrt[n!]{2})$, whenever $1\le k\le n$, so
$$
u\in \mathbb{Q}(\sqrt[m!]{2})
$$
and so $u$ is algebraic over $\mathbb{Q}$, having degree at most $m!$ (prove it).

The extension is not finite, because $\sqrt[n]{2}$ has degree $n$ over $\mathbb{Q}$, the polynomial $X^n-2$ being irreducible (prove it).

If $K$ is a finite extension of $\mathbb{Q}$, say $[K:\mathbb{Q}]=k$, then every element of $K$ is algebraic over $\mathbb{Q}$ of degree at most $k$ (prove it, it's the same as before).