If a graph has no isolated or pendant vertices then it contains at least one simple circuit I am trying to prove that if a finite graph has no isolated or pendant vertices then it contains at least one simple circuit.
Let the graph with no isolated or pendant vertices be $(V,E)$. A path in the graph cannot exceed $|V|-1$ since a path of length $m$ passes through $m+1$ vertices. I do not know where to go from here.

Different hint: Every vertex has degree at least two, so the sum of the degrees of the vertices is at least $2|V|$. Being careful about non-connected graphs, show that such a graph cannot be a forest (disjoint collection of trees).
