What to do after this last step?
This sum appears quite often in DSP.

\begin{align}
\sum_{n=0}^{N-1} \exp(-j\alpha n) &amp;\stackrel{(a)}{=} \frac{1- \exp(-j\alpha N)}{1 - \exp(-j\alpha )}\\
&amp;= \frac{e^{-j\alpha N/2}(e^{+j\alpha N/2}- e^{-j\alpha N/2})}{e^{-j\alpha /2}(e^{+j\alpha /2}- e^{-j\alpha /2})}\\
&amp;\stackrel{(b)}{=} e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\alpha/2)}
\end{align}
where the last step $(b)$ comes from identity $e^{+j\theta}-e^{-j\theta} = 2j\sin(\theta)$ and be careful at $(a)$ as Matt's comment.

Substitute $\alpha=k\omega_0=k\frac{2\pi}{N}$, $\sin(N\alpha/2)=\sin(k\pi)=0$ if $k\in\mathbb{Z}$ hence so is the sum.



Update for $\alpha=0$

It is trivial that $\exp(-j\alpha n) \stackrel{\alpha=0}{=} 1$ then the sum equals $N$.

It can be thought as the case of the final result after $(b)$  when $\alpha \to 0$ then
$$e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \stackrel{\alpha \to 0}{\to} N$$
where the limit is evaluated by L'Hôpital's rule.



Further interesting discussion can be found in Laurent Duval's answer.