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It is well-known that the number of acyclic orientations of $K_n$ is $n!$. Does anybody know of a combinatorial argument for this fact which uses the identity: $$n!=\sum_{k=1}^ns(n,k),$$ where the $s(n,k)$ are Stirling numbers of the first kind? If such a thing exists; what do the different Stirling numbers correspond to exactly?

I would also be interested in any other information or references linking Stirling numbers with acyclic orientations. And failing all this, if anybody knows of any other nice combinatorial arguments (that is, not involving the evaluation of the chromatic polynomial at $-1$) for the numbers of acyclic orientations of complete graphs then I would be interested to hear them.

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An acyclic ordering on a complete graph is just the same think as a total order on the vertices. Edges point from small to big. There are clearly n! of these. So it seems you just want to interpret your identity in terms of linear orders. – Benjamin Steinberg Jun 25 '12 at 16:49
More interesting is the generalization to any graph, due to Greene and Zaslavsky. See Corollary 7.4 of their paper at – Richard Stanley Jun 25 '12 at 23:09

hmmm...seems I was being lazy. Did that thing where you get up to do something else and the answer suddenly seems obvious. You just need to label the vertices of each acyclic orientation according to the order it dictates, and choose one to be your "identity". The others are clearly then just permutations of the labels with cycle numbers corresponding to the k in each s(n,k). Thanks for your help Benjamin.

Writing this as an answer so it can be marked as resolved (if I can accept my own answer!).

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