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The problem was that your ultimate conclusion was wrong. It is not in fact the case that "mathematical induction should start with the second step, not the first." It's just that, like all proofs, you have to draw valid inferences at each step. As JGWeissman points out, the horse proof fails at the n=2 step. But one could contrive examples in which the induction proof fails at the kth step for arbitrary k.
I don't think I ever got to my "ultimate" conclusion (that all of the operations that appear in step n must appear in the basis step).
I was trying to use this example where the proof failed at n=2 to show that it's possible in principle for a (specific other) proof to fail at n=2. Higher-order basis steps would be necessary only if there were even more operations.