Using Euler's formula in graph theory where
$r - e + v = 2$
I can simply do induction on the edges where the base case is a single edge and the result will be 2 vertices. A single edge also has only one region which is the external region.
$r - 1 + v = 2$
$1 - 1 + v = 2$
$v = 2$
Which is true if I drew a connected graph of that has a single edge there are two vertices. However, the inductive step where I am supposed to prove that there are $e + 1$ edges I am confused on how to prove this. I can simply add another edge and the formula will still stand true but I don't think that's the correct way to do it. For example, I don't think my professor would accept this on the exam if I tried doing the inductive step like this:
$r - e + v = 2$
$1 - (e + 1) + v = 2$
if we let $e=1$ again then there would be
$1 - 2 + v = 2$
$v = 3$
which is true, a connected graph of two edges has 3 vertices at most. However, i do not think this is the correct way to prove Euler's formula during the inductive step.