I am starting to self study Galois Theory from J.S. Milne's notes on the subject. He claims that
If $\alpha$ is a constructible number then $\alpha$ is algebraic over $\mathbb{Q}$ and $[\mathbb{Q}[\alpha],\mathbb{Q}]$ is a power of $2$.
Now to prove this he uses 2 facts that are
$1)$ If $F\subset E\subset L$ are fields then $[L:F]=[L:E][E:F]$
$2)$ A number is contructible if and only if it is contained in a subfield of $\mathbb{R}$ of the form $\mathbb{Q}[\sqrt{a_1},...,\sqrt{a_n}]$ with $a_i \in Q[\sqrt{a_1},...,\sqrt{a_{n-1}}]$ and $a_i>0$.
Now to prove the result he just claims that $[\mathbb{Q}[\alpha],\mathbb{Q}]$ divides $[\mathbb{Q}[\sqrt{a_1},...,\sqrt{a_n}],\mathbb{Q}]$ and the latter is a power of $2$. Now I have some doubts about this , I get that since the number is algebraic we will have that $\alpha \in \mathbb{Q}[\sqrt{a_1},...,\sqrt{a_n}]$ and so we will have that $\mathbb{Q}[\alpha]$ will also be there and then we can use the formula , what I don't get is how do we know that $[\mathbb{Q}[\sqrt{a_1},...,\sqrt{a_n}],\mathbb{Q}]$ is a power of $2$? If I have this I get both the results but I don't even know how this would be finite since I don't know that $\sqrt{a_n}$ are algebraic right ?
Maybe I am confusing things up but any help is aprecciated. Thanks in advance.
$\sqrt{a_n}$ is a root of $X^2-a_n$ and thus algebraic of degree 1 or 2.
Edit. I will add details. You have a tower $\mathbb{Q}\subset ...\subset\mathbb{Q}[\sqrt{a_1},...,\sqrt{a_{i-1}}]\subset \mathbb{Q}[\sqrt{a_1},...,\sqrt{a_i}]\subset ...$ and each step is of degree 1 or 2, since $a_i \in Q[\sqrt{a_1},...,\sqrt{a_{i-1}}]$, so the total degree will be of the form $1^a2^b$.