You can start in this way:
$$\alpha = e^{\displaystyle i\frac{2a\pi}{m}}, \beta = e^{\displaystyle i\frac{2b\pi}{n}},$$
with $a, b \in \mathbb{Z}$.
Also, notice that:
$$\begin{align}
\alpha + \beta & = e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)}\left(e^{ i\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)}+e^{ -i\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)}\right) \\
& = 2e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)}\cos\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)
\end{align},$$
and then:
$$(\alpha + \beta)^{nm} = 2^{nm}e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)nm}\left(\cos\left(\frac{a\pi}{m}- \frac{b\pi}{n}\right)\right)^{nm}.$$
We can forget about $2\cos(\ldots)$, it is real!
Moreover...
$$e^{ i\left(\frac{a\pi}{m}+ \frac{b\pi}{n}\right)nm} = e^{ i(an + bm)\pi}.$$
Since $an+bm$ is integer, then $ e^{ i(an + bm)\pi} = \pm 1$.