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.
 A: The good news is that looks like you have all the pieces of information required; you just need to arrange them into the correct order to make the argument.  
As you noted, if $f$ is an irreducible polynomial of degree $n$ over some field $F$, then $F(\alpha)$ is a degree $n$ field extension of $F$, for any root $\alpha$ of $f$.  This fact, together with the fact that $f(x) = x^n - 2$ is irreducible over $\mathbb{Q}$ per Eisenstein gives you what you want.  
In particular, suppose the extension in question is finite, say of degree $m$, and consider $\sqrt[n]{2}$ for some $n>m$.  Since $\sqrt[n]{2} \in E$ by assumption, we must have a tower of inclusions $\mathbb{Q} \subset \mathbb{Q}( \sqrt[n]{2}) \subset E$: a contradiction per the multiplicativity formula. 

P.S.: Be careful with this statement:

let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$.

This is not quite true.  The statement you're looking for is that $u$ can be written as some combination of powers of elements from a finite subset of the generating set.  To be specific, the notation $F(S)$ for some set $S$ denotes $\Big\{f(u_1, u_2, ..., u_k) \ \Big| \ u_i's \in S, \ f \in F[x_1, x_2, ..., x_k], \ k \in \mathbb{N} \Big\}$.  For a concrete example, notice that $\displaystyle 1 + 5\sqrt[3]{2} - \frac{4}{7}(\sqrt[3]{2})^2 \in \mathbb{Q}(\sqrt[3]{2})$.  The conclusion is salvageable, however, since it can be shown that the set of algebraic numbers is a field (hence addition/multiplication of algebraic numbers yields an algebraic number).  In this case, $S = \{ \sqrt[k]{2} \ | \ k \in \mathbb{N} \}$ is comprised entirely of algebraic elements, so $E$ is algebraic.
A: 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 $\sqrt[r+1]2\notin E$.
A: 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).
A: For every $n\in\Bbb N^*$ the field $\Bbb Q(\sqrt[n]{2})$ is a $\Bbb Q-$ vector space of dimension $n$.
Because $[\Bbb Q(\sqrt[n]{2}):\Bbb Q]=n$ is also the dimension of $\Bbb Q(\sqrt[n]{2})$ over $\Bbb Q$.
Now: if $m,n$ are coprime natural numbers, then $\Bbb Q(\sqrt[n]{2},\sqrt[m]{2})$ is a $\Bbb Q$ vector space of dimension $m+n-1$.
Finally: let $F=\Bbb Q(\{\sqrt[p]{2}: p \text{ is prime }\})$ the subfield of $E$. This is also a vector subspace of $E$ and a vector space over $\Bbb Q$ of infinite dimension by the previous discussion. Thus $E$ has infinite dimension.
