I'm trying to make sense of this proof here https://math.stackexchange.com/a/94488/737935
If it's a slick proof you want, nothing beats this proof that the only cases where both $r$ and $\cos(r \pi)$ are rational are where $\cos(r \pi)$ is $-1$, $-1/2$, $0$, $1/2$ or $1$. If $r=m/n$ is rational, $e^{i \pi r}$ and $e^{-i \pi r}$ are roots of $z^{2n} - 1$, so they are algebraic integers. Therefore $2 \cos(r \pi) = e^{i \pi r} + e^{-i \pi r}$ is an algebraic integer. But the only algebraic integers that are rational numbers are the ordinary integers. So $2 \cos(r \pi)$ must be an integer, and of course the only integers in the interval $[-2,2]$ are $-2,-1,0,1,2$.
I cannot find any reason to justify $e^{i \pi r}$ and $e^{-i \pi r}$ being algebraic integers. There are roots of unity in this form that are not algebraic integers, such as $e^{\frac{4\pi i}{3}}$, which is a root of $z^{2*3}=1$.
