This was a proposed exercise in some lecture notes on differential topology I’m reading. Some proofs for this result are given in the answers to this question. However, they assume previous knowledge of algebraic topology, which the lecture notes don’t. That’s why I was curious to know if there’s an alternative proof.
The definition of degree of a map I’m using is the following: Let $f:X \rightarrow Y$ be a smooth map between two oriented manifolds of the same dimension. Let $y\in Y$. Then $\deg_y(f)=I(f,\{y\})$, where $I(f,\{y\})$ is the intersection number of $f$ with $\{y\}$. If we also have that $f^{-1}(y)$ is a finite set $\forall y \in Y$ and that $Y$ is a connected manifold, then we define $\deg(f)=\deg_y(f)$ for any $y \in Y$.
I know as well that if $f,g:X \rightarrow Y$ are smooth maps such that $X$ is a compact manifold, $Y$ is a connected manifold and $f$ is homotopic to $g$, then $\deg(f)=\deg(g)$. I was thinking we could maybe use this in addition to the fact that the map $z\rightarrow z^n$ is homotopic to the identity map concatenated with itself $n$ times to prove that $\deg(f)=n$ (where $f: S^1 \to S^1$ is such that $f(z) = z^n$)?
I'll turn my comments into an answer.
You can apply the definition of degree directly. Starting with the map $f(z)=z^n$, and using $y = 1 + 0i \in S^1$, the set $f^{-1}(y)$ consists of $n$ equally spaced points around $S^1$, namely the $n^{\text{th}}$ roots of unity, $x = \exp(2 \pi i k/n)$, $k=0,...,n-1$.
So, for each $x \in f^{-1}(y)$, you just have to prove that the local degree of $f$ at $x$ is equal to $+1$, because then the global degree is $I(f,\{y\}) = n \cdot 1 = n$. In fact it's just as easy to ignore the special value $y=1+0i$, and prove this local degree fact for all $x \in S^1$ and $y=f(x) \in S^1$.
What this local degree computation comes down to is to prove that the map $$df_x : T_x S^1 \to T_{f(x)}S^1 = T_y S^1 $$ takes the standard orientation on $T_x S^1$ to the standard orientation on $T_y S^1$.