From any vertex we have n−1 edges to choose from first vertex, n−2 edges to choose from second vertex, n−3 to choose from the third and so on. These being independent choices, we get (n−1)! possible number of choices.
Each Hamiltonian circuit has been counted twice (in reverse direction of each other like these: A→B→C→A and A→C→B→A).
Number of distinct not edge disjoint Hamiltonian circuits in complete graph Kn is (n−1)!/2