real analysis - 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?

Answer

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

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

And we are done!