[proofplan]
We formalize the classical argument that every constructible real number is algebraic over $\mathbb{Q}$ with degree a power of $2$. The proof proceeds in three stages: (1) model each ruler-and-compass step as a field extension of degree $1$ or $2$; (2) apply the tower law to show the cumulative extension has degree $2^s$; (3) use the multiplicativity of degrees to conclude that $[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $2^s$, hence is itself a power of $2$.
[/proofplan]
[step:Setting up the tower of extensions]
A point in the plane is constructible if it can be obtained from the two initial points $0$ and $1$ by a finite sequence of ruler-and-compass operations. Each such operation—intersecting two lines, a line and a circle, or two circles—produces coordinates that satisfy at most a quadratic equation over the field generated by the previously constructed coordinates.
Set $K_0 = \mathbb{Q}$. At the $i$-th step we adjoin the new coordinates obtained from that operation, producing a field $K_i$. The new coordinates satisfy either a linear or quadratic equation with coefficients in $K_{i-1}$, so
\begin{align*}
[K_i : K_{i-1}] \in \{1, 2\}.
\end{align*}
After $n$ steps we have a tower
\begin{align*}
\mathbb{Q} = K_0 \subseteq K_1 \subseteq K_2 \subseteq \cdots \subseteq K_n.
\end{align*}
[/step]
[guided]
The degree is $1$ when the new point already lies in $K_{i-1}$ (e.g., the intersection of two non-parallel lines with coefficients in $K_{i-1}$ yields coordinates in $K_{i-1}$ itself). The degree is $2$ when we must adjoin a square root—this occurs whenever a line or circle meets a circle, producing coordinates that satisfy an irreducible quadratic over $K_{i-1}$.
[/guided]
[step:Applying the tower law]
By the multiplicativity of field extension degrees,
\begin{align*}
[K_n : \mathbb{Q}] = [K_n : K_{n-1}] \cdot [K_{n-1} : K_{n-2}] \cdots [K_1 : K_0].
\end{align*}
Each factor is $1$ or $2$, so there exists an integer $s$ with $0 \leq s \leq n$ such that
\begin{align*}
[K_n : \mathbb{Q}] = 2^s.
\end{align*}
[/step]
[step:Concluding the degree constraint on $\mathbb{Q}(\alpha)$]
Let $\alpha \in \mathbb{R}$ be constructible. Then $\alpha$ is a coordinate of some constructed point, so $\alpha \in K_n$ for some tower as above. In particular $\alpha$ is algebraic over $\mathbb{Q}$, since $[K_n:\mathbb{Q}] = 2^s$ is finite and every element of a finite extension is algebraic.
Now $\mathbb{Q}(\alpha) \subseteq K_n$, and the tower law gives
\begin{align*}
[K_n : \mathbb{Q}] = [K_n : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}].
\end{align*}
Therefore $[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $[K_n:\mathbb{Q}] = 2^s$. Since every divisor of $2^s$ is a power of $2$, we conclude
\begin{align*}
[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2^k
\end{align*}
for some integer $0 \leq k \leq s$. $\blacksquare$
[/step]
[guided]
The final divisibility argument is the key point: we do not claim that $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^s$, only that it divides $2^s$. The minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $2^k$ for some $k \leq s$, which is the sharpest conclusion the tower method yields. This is sufficient, for instance, to prove that $\sqrt[3]{2}$ is not constructible: its minimal polynomial $x^3 - 2$ has degree $3$, which is not a power of $2$.
[/guided]
