This was done in Hatcher's algebraic topology example 2.32. However I do not understand it at all.
An alternative approach I know of is to show that the degree for homology matches the degree for the fundamental group which requires Hurwitz map.
For Hatcher's proof, I know that the map is locally a homeomorphism because by inverse function theorem. Then I need to determine the sign of it. How to do this mathematically instead of by words?
Does the sign have anything to do with whether the map is orientation-preserving or not?
If you know that $H^1_{\mathrm{dR}}(S^1)\cong H^1(S^1;\mathbb{R})$, then using the generator $d\theta$ of de Rham cohomology, for $f(\theta)=n\theta$ (on $S^1$ parameterized with $\theta$ modulo $2\pi$) we have $df=n\, d\theta$, hence the induced map $$f^*:H^1_{\mathrm{dR}}(S^1)\to H^1_{\mathrm{dR}}(S^1)$$ is multiplication by $n$. The de Rham theorem is that there is a natural isomorphism to singular cohomology, so paired with the naturality of the universal coefficient theorem, there is the following commutative diagram:$\require{AMScd}$ \begin{CD} H^1_{\mathrm{dR}}(S^1) @>\cong>> H^1(S^1;\mathbb{R}) @>\cong>> \hom(H_1(S^1),\mathbb{R})\\ @VVf^*V @VVf^*V @VV(f_*)^*V\\ H^1_{\mathrm{dR}}(S^1) @>\cong>> H^1(S^1;\mathbb{R}) @>\cong>> \hom(H_1(S^1),\mathbb{R})\\ \end{CD} The UCT isomorphisms come from the fact that the $\operatorname{Ext}^1$ groups are trivial. We know that $H_1(S^1)=\mathbb{Z}$ so $\hom(H_1(S^1),\mathbb{R})=\mathbb{R}$ and $(f_*)^*$ is multiplication by $n$. This implies $f_*$ itself is multiplication by $n$, and therefore $\deg f=n$.
The Hurwitz map works, too. The horizontal maps in the following are surjections: \begin{CD} \pi_1(S^1) @>>> H_1(S^1) \\ @VVf_*V @VVf_*V \\ \pi_1(S^1) @>>> H_1(S^1) \end{CD} Since $\pi_1(S^1)=\mathbb{Z}$ is already abelian, they are of course isomorphisms. The first $f_*$ is the induced map of the covering map $f:S^1\to S^1$ in the case $n\geq 1$. From covering space theory, we can see the image is $n\mathbb{Z}$ with $1\mapsto n$, hence $\deg f=n$. For $n=0$, this is a constant map and $\deg f=0$. For negative $n$, by composition with $z\mapsto z^{-1}$ (which is degree-$(-1)$, mentioned in Hatcher as property (e) of that chapter), we get a covering map, and by property (d) we get $\deg f=(-1)(-n)=n$. There are other ways of being careful here.
The Hatcher approach, Proposition 2.30, is to let $x_1,\dots,x_m$ be the preimage of a point through $z\mapsto z^n$, with $m=\lvert n\rvert$ (and assume $n\neq 0$ since constant functions have already been handled). Concretely, the preimage of $1$ is the $m$th roots of unity, $1,e^{2\pi i/m}, e^{2\pi i\cdot 2/m}, \dots, e^{2\pi i(m-1)/m}$. Using $\theta$ coordinates, each has a an open neighborhood $U_k=(\pi (2k-1)/m, \pi (2k+1)/m)$ for $k=1,2,\dots,m$, and these are disjoint. The images of these through $z\mapsto z^n$ map such a neighborhood onto $(-\pi,\pi)$, possibly reversing orientation if $n$ is negative. The group $H_1(U_k,U_k-e^{2\pi ik/m})$ is $\mathbb{Z}$, generated by a singular $1$-simplex that crosses from one component of punctured $U_k$ to the other. The induced map $$f_*:H_1(U_k,U_k-e^{2\pi ik/m})\to H_1((-\pi,\pi),(-\pi,\pi)-0)$$ either sends this generator to a generator in the same direction, or in the opposite direction, depending on the sign of $n$ (and if you want to be careful here, consider the naturality of the long exact sequence of both pairs). Hence the local degree is $\deg f|e^{2\pi ik/m}=\operatorname{sgn} n$. Applying the proposition, $$\deg f=\sum_{k=1}^m \deg|e^{2\pi ik/m} = \sum_{k=1}^m\operatorname{sgn} n=n.$$