This may qualify, though there is a special case hidden inside the argument.

Consider a simplified game of nim in which there are $n>0$ matchsticks, a player may remove $1$, $2$, or $3$ matchsticks each turn, and the player who takes the last matchstick wins.

Theorem. The first player has a winning strategy if $n\not\equiv 0\pmod{4}$; the second player has a winning strategy if $n\equiv 0\pmod{4}$.

Proof. Strong induction on $n$. Assume the result holds for all $k\lt n$. If $n\equiv 0\pmod{4}$, then after player 1's turn there will be $k\lt n$ matchsticks left, with $k\not\equiv 0 \pmod{4}$. By the induction hypothesis, the first person to play at this point has a winning strategy, this being player 2; thus, player 2 has a winning strategy.

If $n\not\equiv 0\pmod{4}$, then write $n=4\ell + t$ with $1\leq t\leq 3$. Have player 1 take $t$ matchsticks, leaving $4\ell$ matchsticks. If $\ell=0$, player 1 just won. If $\ell>0$, then there are $k\lt n$ matchsticks left, with $k\equiv 0\pmod{4}$. By the induction hypothesis, the player who moves second has a winning strategy, this being the original player 1. So player 1 has a winning strategy in this case.