Theorem: Every integer $n>1$ is either prime or a product of primes. Proof: By strong induction. Let $n$ be an arbitrary natural number greater than 1. Inductive hypothesis: Assume that for every integer $1<k<n$ that $k$ is either prime or the product of primes. If $n$ is prime then no further work is necessary. If $n$ is not prime, then we can choose some natural numbers $a$ and $b$, such that $n=ab$, where $a<n$ and $b<n$. Thus, by the inductive hypothesis, each of $a$ and $b$ is either prime or a product of primes, and since $n=ab$, it follows that $n$ is a product of primes. $□$
My question: Given only this proof, how do we know that the induction hypothesis is not false? If it was false, then we would not have been able to conclude from it that n must be the product of primes. I understand that that the theorem is true and even that it makes intuitive sense, but given the reason just stated I do not understand why this particular proof can be said to have proved the theorem. Can someone please help me pinpoint the gap in my reasoning?
EDIT: My question is about strong induction in general. The book I am using makes the claim that we do not need to show any base case for a proof by strong induction because it is built into the proof itself. The proof above is an example of the author's understanding of proof by strong induction. If a base case was given in the above proof I think I would have less of an issue with it, but I have a feeling that my question about the assumption not being valid would still linger. Here is a quote from the book that explains why the base case is said to e built in (How to Prove it A structured Approach): "Suppose that we’ve followed the strong induction proof strategy and proven the statement ∀n[(∀k < nP(k)) → P(n)]. Then, plugging in 0 for n, we can conclude that (∀k < 0P(k)) → P(0). But because there are no natural numbers smaller than 0, the statement ∀k < 0P(k) is vacuously true. Therefore, by modus ponens, P(0) is true. (This explains why the base case doesn’t have to be checked separately in a proof by strong induction; the base case P(0) actually follows from the modified form of the induction step used in strong induction.) Similarly, plugging in 1 for n we can conclude that (∀k < 1P(k)) → P(1). The only natural number smaller than 1 is 0, and we’ve just shown that P(0) is true, so the statement ∀k < 1P(k) is true. Therefore, by modus ponens, P(1) is also true."