Well, there is one set of cardinality 0; one set of cardinality one; one set of cardinality 2, but since its automorphism group has order 2, we only count it with multiplicity 1/2; there is one set of cardinality 3, counted with multiplicity 1/3!; ... So the number of sets is $$1 + 1+ 1/2! + 1/3! + \dots = e$$