Show that $e^{1-n} \leq \frac {n!}{n^n}$

\begin{align*} n^{n}=(1+(n-1))^{n}\leq n!\sum_{k=0}^{\infty}\dfrac{(n-1)^{k}}{k!}=n!\cdot e^{n-1}. \end{align*}

Using induction we see that for $n=1$, the inequality holds. Assume that it holds for some number $k$.

Then, using $\left(1+\frac1k\right)^k<e$, we find that

$$\begin{align} \frac{(k+1)!}{(k+1)^{k+1}}&=\frac{k!}{k^k\left(1+\frac1k\right)^k}\\\\ &\ge \frac{e^{1-k}}{e}\\\\ &=e^{1-(k+1)} \end{align}$$

And we are done!

Hint: $$ \frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}}=\left(1+\frac1n\right)^n\lt e $$