The sum of 1/(1 − ω^k)

A pair. Put $\theta=\frac{2\pi k}n\in(0,2\pi)$ and $u=e^{i\theta}$. Then $\overline u=\omega^{n-k}$ and
$$\frac1{1-u}+\frac1{1-\bar u}=\frac{2-u-\bar u}{(1-u)(1-\bar u)}=\frac{2-2\cos\theta}{2-2\cos\theta}=1.$$

Summing. Let $S=\sum_{k=1}^{n-1}\frac1{1-\omega^k}$. Since $k\mapsto n-k$ permutes $\{1,\dots,n-1\}$, also $S=\sum_{k=1}^{n-1}\frac1{1-\omega^{n-k}}$. Adding the two expressions of $S$,
$$2S=\sum_{k=1}^{n-1}1=n-1.$$

Answer. $S=\dfrac{n-1}2$.

Checks. $n=2$: $\frac1{1-(-1)}=\frac12$ ✓. $n=3$: the two terms are conjugate and add up to $1$ ✓.

Lesson. Symmetry $k\leftrightarrow n-k$ (complex conjugation) pairs terms; the sum of the pair is often constant.