TITLE: Strong induction without a base case
QUESTION [42 upvotes]: Strong induction proves a sequence of statements $P(0)$, $P(1)$, $\ldots$ by proving the implication 
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true." 
for every nonnegative integer $n$.  There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously.  But most strong induction proofs nevertheless seem to involve a separate argument to handle the base case (i.e., to prove the implication for $n=0$).
Can you think of a natural example of a strong induction proof that does not treat the base  case separately?  Ideally it should be a statement at the undergraduate level or below, and it should be a statement for which strong induction works better than ordinary induction or any direct proof.

REPLY [3 votes]: One may transform any strong induction proof into one which (legalistically speaking!) doesn't explicitly treat the base case.
Proving $P(n)$ by induction, one assumes $ \forall k<n \ P(k)$ which weakens to $\forall k<n\ P(0)\implies P(k)$.`  Now one may adapt whatever proof one had for the induction step to establish from this that $P(0)\implies P(n)$.  Now one proves $P(0)$ and concludes $P(n)$ by modus ponens.
Yes, I know, morally the base case still got special treatment, but formally now that happens in the induction step.  Thus finding a formal distinction between theorems that require special treatment for the base case and those that don't seems impossible.
In any case, if you can prove $\forall n\geq 0\ P(n)$ by strong induction, you can prove $\forall n>0 \ P(0)\implies P(n)$ by strong induction with no special treatment for the base case, morally or formally.  So examples abound.