prove that if $\alpha \in \mathbb{Q}(\sqrt{2},\sqrt[4]{2},\sqrt[8]{2},...)$ then $\alpha \in \mathbb{Q}(\sqrt[2^n]{2})$ for some $n$ prove that if $\alpha \in \mathbb{Q}(\sqrt{2},\sqrt[4]{2},\sqrt[8]{2},...)$ then $\alpha \in \mathbb{Q}(\sqrt[2^n]{2})$ for some $n$. I have the solutions which state:

Since $\alpha \in \mathbb{Q}(\sqrt{2},\sqrt[4]{2},\sqrt[8]{2},...)$ then $\alpha$ is some rational expression in $\mathbb{Q}$ and a finite number of the generators $\sqrt[2^i]{2}$. If $\sqrt[2^n]{2}$ is the highest root cocuring, then $\alpha \in \mathbb{Q}(\sqrt[2^n]{2})$.

I don't quite understand this proof. I understand that $\alpha $ is a rational expression in $\mathbb{Q}$, but why is it a finite number of the generators when we have infinitely many square root of twos - similarly why can we choose the highest root occurring if there are infinitely many. I feel as though the proof isn't formal enough and I am having troubles describing it formally.
 A: It is clear that $\Bbb Q(\sqrt[2^n]{2})\subseteq \Bbb Q(\sqrt[2^{n+1}]{2})$ so that your field is just
$$\bigcup_{n=1}^\infty \Bbb Q(\sqrt[2^n]{2})$$
But then by definition of a union of sets $x\in \bigcup A_n\implies x\in A_n$ for some $n$.
A: "Highest Root Occurring" means the largest $n$ such that $\sqrt[2^n]{2}$ is in the expression of $a$.
For why you need only a finite number of generators, this comes down to the definition: in any ring (in our case a field), finitely generated or not, every element is a finite linear combination of generators. That's just the convention we as mathematicians go by. I guess you can read about Hilbert spaces if you want to see what happens if you loosen this restriction, but...
EDIT: another perspective is to look at $\mathbb{Q}(\sqrt{2}, \sqrt[4]{2}, \dots)$ as a vector space over $\mathbb{Q}$ spanned by $\{1, \sqrt{2}, \sqrt[4]{2}, \dots\}$. Once again, vector spaces require finite linear combinations of basis elements.
A: Your field has infinitely many generators, but any particular element $\alpha$ can use only finitely many of those generators.  The reason is that $\alpha$ has to be a rational expression --- a fraction with polynomial functions of the generators in the numerator and denominator --- and any such expression is finite.  
Then the finitely many generators that are actually used in $\alpha$ can be replaced by one of them, because, in the list $\sqrt 2, \sqrt[4]2, \sqrt[8]2, \dots$, each generator is the square of the next one, so any finitely many generators can all be expressed as powers of the last of them.  
Another way to approach the problem, without talking about "expressions", is to use (what should be) the definition of $\mathbb Q(\sqrt 2, \sqrt[4]2, \sqrt[8]2, \dots)$, namely the smallest field that contains $\mathbb Q$ and all the listed generators.  So all you need to show is that $\bigcup_n\mathbb Q(\sqrt[2^n]2)$ is a field.  Since it obviously contains $\mathbb Q$ and all the generators, it will include $\mathbb Q(\sqrt 2, \sqrt[4]2, \sqrt[8]2, \dots)$.  
