At the end of a proof you write Q.E.D, which stands not for
Quod Erat Demonstrandum as the books would have you believe, but
for Quite Easily Done.
 R. Ainsley in Bluff your way in Maths, 1988

Euler's formula: A connected plane graph with n vertices, e edges and f faces
satisfies n  e + f = 2
Proof. Let T be the edge set of a spanning tree for G. It is a subset of the
set E of edges. A spanning tree is a minimal subgraph that connects all the
vertices of G. It contains so no cycle. The dual graph G* of G has a vertex
in the interior of each face. Two vertices of G* are connected by an edge if
the correponding faces have a common boundary edge. G* can have double edges
even if the original graph was simple. Consider the collection T* of edges
E* in G* that correspond to edges in the complement of T in E. The edges of
T* connect all the faces because T does not have a cycle. Also T* does not
contain a cycle, since otherwise, it would seperate some vertices of G
contradicting that T was a spanning subgraph and edges of T and T* don't
intersect. Thus T* is a spanning tree for G*. Clearly e(T)+e(T*)=2.
For every tree, the number of vertices is one larger than the number of
vertices. Applied to the tree T, this yields n = e(T)+1, while for the tree
T* it yields f=e(T*)+1. Adding both equations gives n+f=(e(T)+1)+(e(T*)+1)=e+2.
 from M.Aigner, G. Ziegler "Proofs from THE BOOK"
