I know intuitively how to construct the map in question, namely by $f(e^{i\theta})=e^{im\theta}$. But I'm having trouble with proving that $\deg(f)=m$. You could prove it by showing that the chains $m\cdot\phi\:$ and $f\circ\phi\:$ are homologous if you take $\phi + B_1(S^{1})$ to be a generator of $H_1(S^{1})$. But how do I do this? This would be a rather direct proof but are there some other theorems that would prove it indirectly?
Cheers.
If you’re okay with computing degree as a sum of local degrees, then note that if you take any point in the image, it has $m$ points in the preimage. At each of these points, $f$ is orientation-preserving so all the local degrees are 1 so you get $deg(f) = \sum_{i=1}^m 1 = m$.
Alternatively, if you know explicit generators of $H_1(S^1)$, you can consider the effect of $f$ on it. In particular, if you take the CW-structure of $S^1$ to be two 0-cells and two 1-cells, denoting the 1-cells by $\Delta_1$ and $\Delta_2$ as Hatcher does, then you can easily see that $deg(f) = f_\ast(\Delta_1 - \Delta_2) = m(\Delta_1 - \Delta_2)$ so the degree is $m$.
There are many other ways of showing this, so depending on what your background is, we could provide suitable answers!