Prove $z \to z^m$ has degree $m$. I am hoping to prove this obeying author's intention - following his hint. But I am wondering if I shouldn't employ Euler's Formula, and should use a more primitive method? I also granted my proof below is correct?

Prove $z \to z^m$ has degree $m$. Hint: calculate with local parametrizations derived from the map $\theta \to (\cos \theta, \sin \theta)$ of $\mathbb{R}^1 \to S^1$.

So we want to show $I(f, \{y\}) = m$. Since any $y$ will work, we pick $(0,1)$ for convenience. By Euler's Formula that
$$e^{ix} = \cos x + i \sin x,$$
or equivalently,
$$(e^{ix})^m = e^{i(mx)}=\cos (mx) + i \sin (mx).$$
Since when $\cos (mx) = 0$ implies $\sin(mx) = 1$, we only need to solve for one.
We solve $\cos (mx) = 0$, clearly $mx =\frac{\pi+2k\pi}{2},$ hence
$$x =\frac{\pi}{2m}+\frac{k\pi}{m}.$$
Clearly, $x$ has $m$ solutions, namely, $k = 0,1,\dots, m-1.$
 A: Heuristic Argument: 
You see that this map, as $z$ moves ever so slightly in the anticlockwise direction, then so does $z^m$. So the map preserves orentation at all points and hence the degree is $m$ as the inverse image has cardinality $m$.
Rigourous argument:
Orient $S^1$ by anticlockwise tangent vectors. An anticlockwise tangent vector must be mapped onto an anticlockwise tangent vector by the derivative of the map $z\mapsto z^m$ at any point $z\in S^1$. It must be so, in particular for the $n$ points in the inverse image.
To see why the above happens, a curve going anticlockwise at uniform angular velocity w.r.t. time $t$ in anticlockwise direction, must go to a curve going uniformly at $m$ times that velocity. So just draw the velocity vector (i.e., tangent vector) at the initial point of the curve. It must go that of the image curve. They must be in the same direction, anticlockwise!
So at each point, the map is orientation-preserving so each point contributes a $+1$ and the degree aught to be $m$.
