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.)
 A: 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.
A: We can state the argument in the following way.  Following the general lines of John Joy's comment:


*

*Suppose we had a polygon of $n > 3$ sides that failed to have $180(n-2)$ internal degrees.  (A triangle cannot fail to have $180$ internal degrees, by assumption.)

*From that polygon, by shaving off a wedge as depicted in the diagram above, we can produce a new polygon of $n-1$ sides that fails to have $180(n-3)$ internal degrees.

*There is no immediate contradiction here, but we eventually must get to a point where we can go no further: We end up with a polygon of $4$ sides that fails to have $360$ degrees.  From this, by the above argument, we can produce a polygon of $3$ sides that fails to have $180$ degrees.  But we have decreed that this cannot happen.

*Thus, our original assumption—that we could have a polygon of $n$ sides that fails to have $180(n-2)$ internal degrees—must be in error.

