Proof about sum of convex polygon interior angles I'm working through Richard Hamming's "Methods of Mathematics Applied to Calculus, Probability, and Statistics" on my own.
I'm struggling with this proof (clipped from Google books):

I follow him all the way until the last sentence (starting with "Thus we have...").
Could someone perhaps take a stab at translating/paraphrasing that rather clunky final sentence, which is the punchline?
(Frustrating that he's so careful and deliberate with the rest of the paragraph (e.g., "This forms a triangle.") while he uses an overly complex sentence at the end to reveal the moral of the story.)

This is really just a convoluted way to describe induction on $n$.
In the first part he proves the following: If we have a convex $n$-gon for which the sum of the angles is $S$, then we can construct an $(n - 1)$-gon for which the sum of the angles is $K - 180^\circ$.
Now this $(n - 1)$-gon can again be reduced to an $(n - 2)$-gon for which the sum of the angles is $K - 2 \cdot 180^\circ$. This step can be repeated until we reach a $3$-gon, i.e. a triangle. The sum of its angles must be $K - (n - 3) \cdot 180^\circ$.
If we now assume $K \ne (n - 2) \cdot 180^\circ$, then the sum of the angles in the triangle isn't equal to $(n - 2) \cdot 180^\circ - (n - 3) \cdot 180^\circ = 180^\circ$. But this is a contradiction, so the formula $K = (n - 2) \cdot 180^\circ$ must be true.
