My understanding of proof by induction is that we first established a base case, say $n = 1 $, and then upon establishing that $P(n)\implies P(n+1)$, we can conclude, by applying this fact an infinite number of times, that $P(n), \forall n$, since $1 \implies 2$, $2 \implies 3$, $3 \implies 4$, and so forth.
How can we reconcile this with inductive proofs that establish that $P(n-1) \implies P(n)$? If I took a base case, for example, of $n = 1$, then I'd have $P(0)$, which may not make any sense in the context. Would I need to take an $n = 2$ base case?
If $P(n)$ is defined only when $n\in\mathbb N=\{1,2,3,\ldots\}$, then, of course, we can't say that$$(\forall n\in\mathbb N):P(n-1)\implies P(n);$$it doesn't make sense. But we can say that$$(\forall n\in\mathbb N\setminus\{1\}):P(n-1)\implies P(n),$$which is the same thing as asserting that$$(\forall n\in\mathbb N):P(n)\implies P(n+1).$$