Euler's formula for plane graphs

Base case. If $E$ $= 0$ and the graph is connected, it must consist of a single vertex ($V=1$), and there is exactly one face, the entire unbounded plane ($F=1$). Then $1 - 0 + 1 = 2$ holds.

Inductive step, case 1: an edge lying on a cycle. Suppose the formula holds for every connected plane graph with fewer than $E$ edges, and let the graph have $E$ $\ge 1$ edges. If some edge $e$ lies on a cycle, then removing $e$ keeps the graph connected (the rest of the cycle still connects its two endpoints). Removing $e$ merges the two faces on its two sides into a single face, so the smaller graph has $V$ vertices, $E - 1$ edges and $F - 1$ faces. By the inductive hypothesis, $V - (E-1) + (F-1) = 2$, which simplifies to $V - E + F = 2$.

Inductive step, case 2: no edge lies on a cycle. Then every edge is a bridge, meaning the graph has no cycles at all, i.e. it is a tree. A tree drawn in the plane has exactly one face ($F = 1$, the single unbounded region, since there is no cycle to enclose any bounded region), and a standard fact about trees is that a tree on $V$ vertices has exactly $E = V - 1$ edges. Substituting, $V - E + F = V - (V - 1) + 1 = 2$.

Conclusion. Every case reduces to the formula holding, or reduces (via removing one edge) to a smaller graph where it holds by induction, so $V - E + F = 2$ for every connected plane graph.