Given that $0<r<n$ and r is much smaller than n, show that $\frac{n-r}{n} \approx e^{\frac{-r}{n}}$

Question

I tried to solve this problem; and I looked at the answer. It said that $e^{\frac{-r}{n}} =\sum_{0}^{\infty}{\frac{(\frac{-r}{n})^k}{k!}}\approx 1-\frac{r}{n} $ I can't under stand the summation approximation.

Thank you so much for your reply.