Please prove by induction.

This problem involves directed graphs. A directed graph is a pair (V, E) where V is a finite set of vertices and E An edge (u, v) means that there is an edge from u to v There can be graphs that have an edge from u to v but not from v to u. V V is a set of edges. For this problem, there's nothing you need to know about directed graphs other than the above definition. A directed graph G = (V,E) is said to be densely connected if for every pair of distinct vertices u and v in V, there is either an edge from u to v or from v to u, but not both. Here is an example densely connected graph. A Hamiltonian path in a directed graph is a path (the path must respect the directionality of edges, of course) that visits all vertices of the graph and visits them exactly once. For example, in the above graph, d -b -->a -->e is a Hamiltionian path Prove that every densely connected graph with more than one vertex has a Hamiltonian path