Here's a cute inequality for unsigned Stirling numbers of the first kind:
$$\genfrac[]{0pt}{}{n}{n-k}\leq\frac{n^{2k}}{2^kk!}.$$
I can prove it using induction (with a beautiful application of AM-GM, see below), but is there a combinatorial proof?

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Here's the core of the induction proof:
$$\begin{align\*}
\genfrac[]{0pt}{}{n}{n-k}&=(n-1)\genfrac[]{0pt}{}{n-1}{n-k}+\genfrac[]{0pt}{}{n-1}{n-k-1}\\
 &=(n-1)\genfrac[]{0pt}{}{n-1}{(n-1)-(k-1)}+\genfrac[]{0pt}{}{n-1}{(n-1)-k}\\
 &\leq(n-1)\frac{(n-1)^{2(k-1)}}{2^{k-1}(k-1)!}+\frac{(n-1)^{2k}}{2^kk!}\\
 &=\frac{1}{2^kk!}(2k+n-1)(n-1)^{2k-1}\\
 &\leq\frac{1}{2^kk!}\left(\frac{(2k+n-1)+(2k-1)(n-1)}{2k}\right)^{2k}\\
 &=\frac{n^{2k}}{2^kk!}
\end{align\*}$$
where the last inequality (the penultimate step) uses the AM-GM inequality.
I find it really beautiful how the AM-GM inequality works perfectly here with no further estimates needed.