Euler's Formula, Proof 3: Induction on Vertices

This argument is the planar dual to the proof by induction on faces.

If \(G\) has only one vertex, each edge is a Jordan curve, so there are \(E+1\) faces and \(F+V-E=(E+1)+1-E=2\). Otherwise, choose an edge \(e\) connecting two different vertices of \(G\), and contract it. This decreases both the number of vertices and edges by one, and the result then holds by induction.