Show that $e^{1-n} \leq \frac {n!}{n^n}$ How can I show that for a $n \in \mathbb N$
$$e^{1-n} \leq \frac {n!}{n^n}$$
I tried using the binomial theorem like this
$$n^n \le (1+n)^n = \sum_{k=0}^n \binom nk n^k \le \sum_{k=0}^\infty \binom nk n^k = \sum_{k=0}^\infty \frac{n!}{k!(n-k)!} n^k \le \sum_{k=0}^\infty \frac{n!}{k!} n^k = n! \sum_{k=0}^\infty \frac{n^k}{k!} = n! \cdot e^n$$
which would give me
$$\frac{1}{e^n} \le \frac{n!}{n^n}$$
But I'm missing the factor of $e$ on the left side. Can you give me a hint?

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!
