Proof.
Let \(P(n)\) be the statement, “every connected planar graph containing \(n\) edges satisfies \(v - n + f = 2\text{.}\)” We will show \(P(n)\) is true for all \(n \ge 0\text{.}\)
Base case: there is only one graph with zero edges, namely a single isolated vertex. In this case \(v = 1\text{,}\) \(f = 1\) and \(e = 0\text{,}\) so Euler’s formula holds.
Inductive case: Suppose \(P(k)\) is true for some arbitrary \(k \ge 0\text{.}\) Now consider an arbitrary graph containing \(k+1\) edges (and \(v\) vertices and \(f\) faces). No matter what this graph looks like, we can remove a single edge to get a graph with \(k\) edges which we can apply the inductive hypothesis to.
There are two cases: either the graph contains a cycle or it does not. If the graph contains a cycle, then pick an edge that is part of this cycle, and remove it. This will not disconnect the graph, and will decrease the number of faces by 1 (since the edge was bordering two distinct faces). So by the inductive hypothesis we will have \(v - k + f-1 = 2\text{.}\) Adding the edge back will give \(v - (k+1) + f = 2\) as needed.
If the graph does not contain a cycle, then it is a tree, so has a vertex of degree 1. Then we can pick the edge to remove to be incident to such a degree 1 vertex. In this case, also remove that vertex. The smaller graph will now satisfy \(v-1 - k + f = 2\) by the induction hypothesis (removing the edge and vertex did not reduce the number of faces). Adding the edge and vertex back gives \(v - (k+1) + f = 2\text{,}\) as required.
Therefore, by the principle of mathematical induction, Euler’s formula holds for all planar graphs.