The typical characterization of points constructible by compass and straightedge is the following:


Let $S\subseteq\mathbb{C}$ with $0,1\in S$, $K\_0 = \mathbb{Q}(S\cup \bar{S})$ and $a\in\mathbb{C}$.
Then $a$ is constructible from $S$ by compass and straightedge if and only if there is a tower of quadratic field extensions $K\_0 \subseteq \ldots \subseteq K\_n$ such that $a\in K\_n$.


For constructible $a$ it follows that $a$ is algebraic over $K\_0$ and $[K\_0(a) : K\_0]$ is a power of two.
However, it is known that this is not sufficient for $a$ to be constructible.


Now I wonder if the constructibility of $a$ is equivalent to the following sharper criterion:


$a$ is algebraic over $K\_0$ and the degree of the normal hull of $K\_0(a)$ over $K\_0$ is a power of two.


The direction ,,$\Leftarrow$'' is true, I think.
If $N$ is the normal hull of $K\_0(a)$, then $K\_0\subseteq N$ is a finite Galois extension, and thus the order of $G = \operatorname{Gal}(K\_0 \subseteq N)$ is a power of two.
As a $2$-group, it contains a chain of subgroups $\{\operatorname{id}\} = U\_n < \ldots < U\_0 = G$ of index $2$ each.
The respective fixed fields give the needed tower of quadratic field extensions.


But I wasn't able to proof ,,$\Rightarrow$'', nor did I find a counter example.



Algebraic characterization of points constructible by compass and straightedge
Your question is about showing that the normal hull of $K\_n$ over $K\_0$ has $2$-power degree, if $[K\_i:K\_{i-1}]=2$ for all $i$. But that follows be induction: Let $L$ be the normal hull of $K\_{n-1}$ over $K\_0$, so $[L:K\_0]$ is a $2$-power.


The normal hull $N$ of $K\_n$ over $K\_0$ is the composite of the conjugates of $K\_n$ over $K\_0$. But all these conjugates are extensions of $L$ of degree $2$ (or $1$, if $K\_n\subseteq L$), and from that the claim follows.


