Show that $\binom{n}{k} \frac{1}{n^k}\leqslant \frac{1}{k!}$ holds true for $n\in \mathbb{N}$ and $k=0,1,2, \ldots, n$ $$\binom{n}{k} \frac{1}{n^k}\leqslant \frac{1}{k!}$$
How would I prove this? I tried with induction, with $n$ as a variable and $k$ changing, but then I can't prove for $k+1$, can I? 
Is there a better method than using induction (if induction even works)?

The simplest way I see is the following:
$$\binom{n}{k}\frac{1}{n^k} = \frac{\overbrace{n(n-1)\dots(n-k+1)}^{\text{$k$ positive terms, each no larger than $n$} }}{k!} \frac{1}{n^k} \le \frac{1^k}{k!}$$
