Show a regular n-sided polygon is constructible, using only ruler and compasses, iff the number $\alpha = 2 \cos(2\pi/n)$ is constructible. Let $n > 2$. Show that it is possible to construct a regular n-sided polygon, using only ruler and compasses, if and only if the number $\alpha = 2 \cos\left(\frac{2\pi}{n}\right) = \zeta_n + \zeta_n^{-1}$ is a constructible length, where $\zeta_n = e^{\frac{2\pi i}{n}}$ is the standard choice of primitive nth root of unity.
This is what I've managed to come up with so far:
Suppose a regular n-sided polygon is constructible for some n. This is equivalent to the angle $\frac{2\pi}{n}$ being constructible and as $2\in\mathbb{Q}\subset C$ we have that $\alpha = 2\cos\left(\frac{2\pi}{n}\right)$ is must also be constructible.
Conversely, Suppose that $\alpha = 2 \cos\left(\frac{2\pi}{n}\right) = \zeta_n + \zeta_n^{-1}$ is constructible. Now, $\alpha$ is constructible iff there is a tower of fields 
$$\mathbb{Q} = K_0 \subset K_1 \subset \dots \subset K_n \subset \mathbb{R}$$
with $\alpha \in K_n$ and $\left[K_{j+1}:K_j\right] \leq 2 \space$ for some $\space 0\leq j \geq n-1$. 
We also know that if $\alpha \in C $ then $\left[\mathbb{Q}(\alpha):\mathbb{Q}\right] = 2^r$ for some integer $r\geq0$.
This is as far as I've managed to get. Is my proof for the forward part sufficient and if so how do I finish the converse part? If not then what have I missed out in both parts?
Thank you in advance.
 A: If $\cos(2\pi/n)$ is constructible, then make a unit circle $C$ intersect with the vertical line through $\cos(2\pi/n)$. The point of intersection gives $\zeta_n$ (treating the Euclidean plane as the complex plane).
A: Much easier (as in: without much knowledge about field extensions, arguing only with ruler/compass-constructions and simple algebraic manipulations):


*

*If you can construct a regular $n$-gon, then (because you can find the centre of the $n$-gon as intersection of some bisectors) you can translate and scale it and so construct a regular $n$-gon with centre $0$ and one vertex $1$. Then the "next" and "previous" vertex are $\zeta_n$ and $\zeta_n^{-1}$. The fourth vertex of the parallelogram with vertices $\zeta_n$, $0$, and $\zeta_n^{-1}$ is of course $2\cos\frac{2\pi}n$.

*If you can construct $\alpha$ (i.e., a point on the line through $0$ and $1$ such that its oriented distance from $0$ is $2\cos\frac{2\pi}n$ times the oriented distance of $1$ from $0$), you can construct the point $2$ by reflecting $0$ at $1$, and the line perpendicular to the "real line" through $a$, and the intersction of that line with  circle around $0$ through $2$. The intersection points are $2\zeta_n$ (and $2\zeta_n^{-1}$), and from there you readily obtain a regular $n$-gon.
