It is shown in this question that
$${n \brack n-k}\leq\frac{n^{2k}}{2^kk!}.$$
But a sharper bound seems to be $${n \brack n-k}\leq\frac{n^{k}}{2^k}{n-1 \choose k}.$$
I don't see how to derive this inequality. Any idea?
Hereafter is some numerical evidence: this is a representation of the natural logarithm of $f(n,k)$ as a function of $k$ in the range $1\le k \le n-1$, for $n=30$. The red dots are for $f(n,k)={n \brack n-k}$, the black dots for $f(n,k)=\frac{n^{2k}}{2^kk!}$ and the blue dots for $f(n,k)=\frac{n^{k}}{2^k}{n-1 \choose k}$.
I think this proof by induction works: \begin{align} {n \brack n-k} &= (n-1){n-1 \brack n-k}+{n-1\brack n-k-1} \\&\le (n-1)\frac{(n-1)^{k-1}}{2^{k-1}}\binom{n-2}{k-1}+\frac{(n-1)^k}{2^k}\binom{n-2}{k} \\&=\frac{(n-1)^k}{2^k}\binom{n-1}{k}\left[\frac{2k}{n-1}+\frac{n-k+1}{n-1}\right] \\&=\frac{n^k}{2^k}\binom{n-1}{k}\cdot {\frac{(n-1)^k}{\color{blue}{n^k}}\cdot\frac{n+k-1}{n-1}} \\&\le\frac{n^k}{2^k}\binom{n-1}{k}\cdot {\frac{(n-1)^k}{\color{blue}{(n-1)^k+k(n-1)^{k-1}}}\cdot\frac{n+k-1}{n-1}} \\&=\frac{n^k}{2^k}\binom{n-1}{k} \end{align} To prove that second inequality, expand $n^k=((n-1)+1)^k$ with the binomial theorem, and only keep the first two terms.