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Apr 29, 2017 at 7:21 comment added Filip Nikšić On the other hand, if we forget about $f$'s being injective, this shows $\sum_{j\leq k} j! {n \brace j} {n \choose j} \leq n^k {n \brace k}$. I'm not sure at the moment what the thing on the left is.
Apr 29, 2017 at 7:12 comment added Filip Nikšić It doesn't quite work. Note that when $f$'s are injections, your $g$'s are not just any functions from $[n]$ to $k$-element subsets of $[n]$, they are surjections. So this actually shows a weaker inequality $k! {n \brace k} {n \choose k} \leq n^k {n \brace k}$, which is just ${n \choose k} \leq n^k / k!$.
Apr 29, 2017 at 1:14 history answered Richard Stanley CC BY-SA 3.0