TITLE: Induction proof: $S$ contains powers of 2 and predecessors implies $S={\bf N}$
QUESTION [6 upvotes]: Let $S$ be a subset of $\mathbb{N}$ such that
$$2^k\in S\quad\forall\ k\in\mathbb N$$
and,
$$k\in S\Rightarrow k-1\in S\quad\forall\ k\ge2,k\in\mathbb N$$
Show by induction that $S=\mathbb N$.

REPLY [9 votes]: Hint $\ $ If one views this as a sort of interval (or segment) induction, then it becomes obvious, namely $\Bbb N$ is the only unbounded initial segment of $\Bbb N.$ Below is the simple proof. 
Lemma $\rm\ S\subset \mathbb N\:$ satisfies $\rm\:n+1\in S\:\!\Rightarrow n \in S\,\ $ iff $\rm\,\ S\:$ is an initial segment of $\:\mathbb N$ 
Proof $ $ (hint) $ $ If $\rm\:S\ne \mathbb N\:$ then $\rm\:S = [0,m)\:$ for the least $\rm\:m\not\in S,\:$ since by induction, nonmembership ascends from $\rm\:m\:$ by $\rm\:n\not\in S\:\Rightarrow\:n+1\not\in S\:$ (contrapositive of the hypothesis).
Corollary $ $ If, additionally, $\rm\:S\:$ is unbounded then $\rm\:S = \mathbb N$
In your case, $\rm\,S\,$ is an initial segment by your second hypothesis and the Lemma, and $\rm\: S\ne \{\ \}\:$ is unbounded via the hypothesis $\rm\:n\in \Bbb N\:\Rightarrow\:2^n\in S.$ Therefore, by the corollary, $\rm\, S = \Bbb N.$
Notice how abstracting the matter set-theoretically helps to make clearer the key structure.
See also this closely related question, where it is referred to as Cauchy's method of induction. Sometimes it is called back-and-forth induction. It occurs frequently  in competition problems.