How to show that $\sum_{k=1}^ne^{ik}=\frac{(e^{in}-1)e^i}{e^i-1}$

Question

The question fits in the title: what tool(s) is/are needed to deduce this identity? $$\sum_{k=1}^ne^{ik}=\frac{(e^{in}-1)e^i}{e^i-1}.$$ This is the last step in my solution to another problem; I've tried playing around with the Binomial theorem to no avail and don't see what to try next. A good hint will suffice!

Answer 1

This is just geometric series with $a = e^i$ and $r = e^i$.

Apply the summation formula for a finite geometric series.

Hint: $\sum_{k=0}^{n} ar^k = \displaystyle\frac{a(r^n-1)}{r-1} \text{ for }r>1$

Answer 2

Hint:

$\sum_{k=1}^ne^{ik}=e^i-e^{i(n+1)}+e^i\cdot\sum_{k=1}^ne^{ik}$

Note: This is a particular case of the geometric summation ($\sum a^i$). The same reasoning exposed for $e^i$ is valid for a general $a$