I wanted to ask if anyone knew of a simple way to prove this.
The proof I know: Identify any such character with a continuous, periodic homomorphism $f: \mathbb R \rightarrow S^1 \subset \mathbb C^{\ast}$. Then $f$ has a Fourier expansion $$f(x) = \sum\limits_n c_n e^{2\pi i nx}$$
with
$$c_n = \int\limits_{\mathbb R/\mathbb Z} f(x)e^{-2\pi i nx}dx = \begin{cases} 1 & \textrm{if $f(x) = e^{2\pi i nx}$} \\ 0 &\textrm{else} \end{cases}$$
This shows that if $f$ is not any of the $e^{2\pi i nx}$, it is zero. $\blacksquare$
This proof could be considered circular, because the Fourier expansion of $f$ is essentially using the Peter-Weyl theorem, which says that the characters of a compact topological group form an orthonormal basis for the $L^2$ space. So we are essentially already assuming that $f$ is one of the $e^{2\pi inx}$.
How can this be proved without using Fourier series/circular reasoning?