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.
 A: $$e^k=\sum_{i=0}^\infty \frac{k^i}{i!}>\frac{k^k}{k!}$$
result follows by taking reciprocal
A: Rewrite it as $$e^{k}>\frac{k^k}{k!}$$
And you see that it follows because $$e^{x}=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$, and thus, for any $x\geq 0$, $e^{x}>\frac{x^k}{k!}$. 
A: 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.
A: 
There have already been several answers posted that relied on the Taylor series for $\displaystyle e^x$.  Herein, we take a different approach to establish the bound by using only straightforward arithmetic and Riemann sums. 

Let $f(k)=\frac{k!}{k^k}$.  Then, $\log(f(k))$ is given by 
$$\begin{align}
\log(f(k))&=\log(k!)-\log(k^k)\\\\
&=\sum_{j=1}^k\log(j)-k\log(k)\\\\
&=\sum_{j=1}^k\log(j/k)\\\\
&=k\underbrace{\left(\frac1k \sum_{j=1}^k\log(j/k)\right)}_{\text{Right-Sided Riemann Sum for}\,\int_0^1\log(x)\,dx=-1}\\\\
&>-k
\end{align}$$
Hence, we have established that $f(k)>e^{-k}$ as was to be shown!
