# Applying fourier transform twice (DSP course)

I stumbled upon a question in a DSP course (coursera) which I don't understand, shown below is a screenshot of the question and answer.

The part which I don't understand is circled. Why is it equal to 0? For multiples of N I understand that each element in the summation is 1 because the exponent is 0. But why does the summation equal 0 when (i+n) is not a multiple of N?

$$\sum_{k=0}^{N-1}e^{-j\frac{2\pi}{N}kl}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}},\qquad l\neq mN,\;m\in\mathbb{Z}$$
Now realize that $$e^{-j2\pi l}=1$$ for $$l\in\mathbb{Z}$$ and the result follows.