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)?

HINT:
Write out the expression 
$$\begin{align}
\binom{n}{k}\frac{1}{n^k}&=\frac{n(n-1)(n-2)\cdots (n-k-2)(n-k-1)}{k!n^k}\\\\
&=\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-2}{n}\right)\left(1-\frac{k-1}{n}\right)
\end{align}$$
