If $p$ is prime and $e^{2\pi i/p}$ is constructible, then $p-1$ must be a power of $2$, since the degree of the irreducible polynomial for $e^{2\pi i/p}$ (which is $\frac{x^p-1}{x-1}$, of degree $p-1$) must be a power of $2$, so $p$ must be a Fermat prime.
Conversely, however, it isn't obvious to me why $e^{2\pi i/p}$ is constructible for every such $p$. In order for this to be the case, $e^{2\pi i/p}$ not only has to have the right degree, but it also has to be expressible in terms of addition, subtraction, multiplication, division, and square roots over $\Bbb{Q}$ (if I'm not mistaken). This isn't always the case with algebraic numbers of degree greater than $4$, and I don't see why it's the case for these roots of unity.
Thank you in advance.
If $p$ is a prime of the form $2^k+1$ and $\omega=\exp\left(\frac{2\pi i}{p}\right)$, the Galois group of $\omega$ over $\mathbb{Q}$ is a cyclic group with order $2^k$, hence the splitting field of its minimal polynomial can be realized by starting from $\mathbb{Q}$ through a chain of extensions of degree $2$. Each extension is associated with the square root of an element in the previous field, hence the extension $\mathbb{Q}(\omega):\mathbb{Q}$ is not only a Galois extension, but a constructible extension in the euclidean sense.