if $a \in \mathbb{Q}(2^{1/2},2^{1/4},2^{1/8},...)$ if $a \in \mathbb{Q}(2^{1/2},2^{1/4},2^{1/8},...)$ then why does it mean that $a$ can be expressed as a finite many different roots of $2$?
The key is to think about how you define something like $\mathbb{Q}(2^{1/2},2^{1/4},\ldots)$.

Defining that is sort of like defining $K[x_1,x_2,x_3,\ldots]$. Even though you've adjoined infinitely many variables, what does an element of $K[x_1,x_2,x_3,\ldots]$ look like? Well, it's a polynomial ring, so every element is a polynomial in arbitrarily many (but finitely so) variables (and with only finitely many terms!)

After all, $x_1 + x_2 + x_3 + x_4 + \cdots$ is not a polynomial.

The same is true for adjoining algebraic elements to a field, since after all the way you define a field extension of $K$ is by taking $K[x]/f(x)$, where $f(x)$ is an irreducible. Thus, your field is (by definition):

$$\mathbb{Q}[x_1,x_2,x_3,x_4,\ldots]/(x_1^2-2,x_2^4-2,x_3^8-2,x_4^{16}-2,\ldots)$$

Where modding out by the ideal $(x_1^2-2,x_2^4-2,x_3^8-2,x_4^{16}-2,\ldots)$ means that you should treat each $x_i$ as $\sqrt[2^i]{2}$.
Your field can be represented as a sequence of successive quadratic extensions $$\mathbb Q\subset\mathbb Q(\sqrt2)\subset\mathbb Q(\sqrt[4]2)\subset\mathbb Q(\sqrt[8]2)\subset\cdots\subset\mathbb Q(\sqrt[2^i]2)\cdots$$ from which you can see that $a$ given, necessarily there is a suitable $2^i$ so the element $a$ belongs to the subfield $\mathbb Q(\sqrt[2^i]2$) of finite degree.

Which is interesting to mention is that  this field is the only one where all construction with ruler and compass  is feasible. For instance the classic impossible problem of the trisection of the angle gives place to a cubic field (because of  the formula for third of the angle) which is not a subfield of your field.

Edit: I understand the downvote: I referred to the kind of field and number 2 obviously does not interfere in what I meant. It  was a distraction for me, it's true.