"Fake proof" of discrete Fourier transform of $x(n)$?

Question

I originally thought about posting this in the Digital Signal Processing StackExchange, but I feel it's more adequate here. Basically, I want to rigorously justify a step which may or not be false - I don't know!

Let $x(n)$ be the indicator function for $0\leq n \leq N-1$ (in signal processing terms, $x(n)=u(n)-u(n-N)$). Then its discrete Fourier transform of length $N$ is given by

$$\sum_{n=0}^{N-1}e^{-2\pi i \frac{n}{N}k}$$

for $0\leq k \leq N-1$. This looks a lot like a geometric series (if $k\neq 0$), hence I feel like I can claim

$$e^{-2\pi i \frac{n}{N}k} = ({e^{-2\pi i \frac{1}{N}k}})^{n}$$

and do the sum as usual. However, for complex exponents isn't this identity that I used false in general? For $k\neq 0$ it gives the correct result, zero.

Answer 1

This is fine, since for real $x$ and integer $n$, we always have $e^{inx} = \left(e^{ix}\right)^{n}$. See De Moivre's formula.