 Solution

The topological group $S^1=\{z\in \mathbb{C}:|z|=1\}$ has complex multiplication and the subspace topology. It is compact and abelian, so every finite-dimensional irreducible complex representation is one-dimensional. A continuous character lifts along $t\mapsto e^{it}$ to a continuous homomorphism $\mathbb{R}\to {\mathbb{C}}^{\times }$, hence has the form $t\mapsto e^{i\lambda t}$. Periodicity by $2\pi$ forces $\lambda =n\in \mathbb{Z}$, giving precisely $z\mapsto z^n$.