I have some difficulties in understanding the proof (written above and taken from here, pp. $5$-$6$) to the following theorem.
Theorem (Euler's formula):
If $G$ is a connected plane graph with $V$ vertices, $E$ edges, and $F$ faces, then $V+F-E=2$.
Proof (by induction on $F$):
$G$ is connected and has only one face. It is a tree, so $E=V-1$ and therefore $V+F-E=V+1-(V-1)=2$.
Now, suppose the formula holds for a connected graph with $n$ faces. We prove that it holds for $n+1$ faces.
Choose an edge connecting two faces of $G$ and remove it. The resulting graph remains connected. The new graph has one fewer edge and one fewer face. So, by the inductive hypothesis, $V+(F-1)-(E-1)=V+F-E=2$.
In the proof above, it's not clear to me how we can "legally" consider a graph with fewer edges and faces. I know this is how sometimes induction works, since you want to reduce it from the "$n+1$" to the "$n$" case, in order to apply the induction hypothesis. But it's just not clicking with me right now.