Why is the following inequality true: $\frac{k!}{k^k} > e^{-k}$ I stumbled upon the following inequality in a scientific paper which estimates a lower bound for $\frac{k!}{k^k}$ for $k \in \mathbb{N}$: 
$$\frac{k!}{k^k} > e^{-k}$$
They did not explain why this holds true, and I could not find any answer by myself yet.

Use the Taylor series:
$$e^k = 1+k+\frac{k^2}{2!} +\cdots+\frac{k^k}{k!} +\cdots.$$
Because all terms on the right are positive, we have
$$e^k > \frac{k^k}{k!},$$
then just take reciprocals of both sides.
