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?

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