Proof of series expansion of $\operatorname{arctanh}(e^{is})$ for $s\in(0,\pi)$.

Question

I would to know how to prove that the series expansion of $\operatorname{arctanh}(e^{is})$ for $s\in(0,\pi)$ is:

$$\sum_{m=1}^{+\infty}\dfrac{e^{is(2m-1)}}{2m-1}.$$

We know by Taylor series that $$\operatorname{arctanh}(z)=\sum_{m=1}^{+\infty}\dfrac{z^{2m-1}}{2m-1},$$ for $|z|<1$. The case $z=1$ is clear, but I don't know how to see the case $z=e^{is}.$

Thanks for all.