The proof as laid out is correct for an arbitrary N. The induction step is to show that it is also true for N+1
Appart for the major woosh, as you didn't get the obvious joke "this holds for any N" -> "no !!! only for N-2" (and I'm not sure at this point that you will even get the hint)
You seem to have a major problem understanding the induction process which you claim to be your prefered and most intuitive way of understanding mathematical proofs. (but since you're a nice person you still admit that your GP's post is good enough for slashdot standards (thank you very much for him/her and the rest of us))
So "The proof as laid out is correct for an arbitrary N" as you said... Why in the world would you need to show that it is also true for N+1 ? N+1 *is* an arbitrary N
Induction is about having specific working examples (not arbitrary !!!) and proving that from here the next candidates must also be valid.
Also, induction is not just about having N and proving N+1, or I could 'prove' you many funny things (sometimes you need more than just one element to get the next and depend on several ones (P(0) and P(1) are true, thus P(2) (depending on the previous two) is true)
Keep the number of passes in a compiler to a minimum. -- D. Gries