Applying fourier transform twice (DSP course)
Just use the formula for the geometric sum:
$$\sum_{k=0}^{N-1}e^{-j\frac{2\pi}{N}kl}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}},\qquad l\neq mN,\;m\in\mathbb{Z}$$
Now realize that $e^{-j2\pi l}=1$ for $l\in\mathbb{Z}$ and the result follows.