Another upper bound for the Stirling numbers of the first kind 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.
