If f is an arbitrary surjection from N onto M, then we can think of f as partitioning N into m different groups, each group of inputs representing the same output point in M. The Stirling Numbers of the second kind count how many ways to partition an N element set into m groups. But this undercounts it, because any permutation of those m groups defines a different surjection but gets counted the same. There are m! such permutations, so our total number of surjections is

m! S(n,m)