A stack of banana-disks, biggest at the bottom, on the left peg. Move it to the right peg — one disk at a time, never a bigger banana on a smaller one. The catch every child meets: each disk you add doubles the work. Slide the height and watch the minimum-move count go 2ⁿ−1, and the solver actually walk it.
The optimal solution is forced by one recursion: to move n disks A→C you must first move the top n−1 out of the way (A→B), move the biggest (A→C), then move the n−1 back (B→C) — so moves(n)=2·moves(n−1)+1, which unrolls to exactly 2ⁿ−1. The solver here IS that recursion; a fail-loud self-check runs it for n=1..10 and throws unless the move-count equals 2ⁿ−1 AND a live peg-simulator confirms every single move is legal (never a larger disk onto a smaller one) and the whole stack lands on peg C.
2ⁿ−1 is the exact optimum for the 3-peg puzzle; with 4+ pegs (the Frame-Stewart problem) the minimum drops and is still not fully proven in general. The banana is decoration; the mechanism is exponential recursion / the Tower of Hanoi (Édouard Lucas, 1883).