Why this is O(n!)
Why O(n!). There are n! permutations of n elements and we produce all of them. The recursion's branching factor decreases by 1 at each level — n × (n−1) × (n−2) × … × 1 = n! leaves.
Output-bound. Like the power-set case, this is a *lower bound*: you can't produce n! results in less than n! work, no matter how clever the generator.
How to recognize factorial complexity in the wild
O(n!) is permutation territory. Generating every ordering, brute-force travelling salesman, anything that asks 'in how many orders can I arrange n things.' Even n=12 is already a billion permutations.
The eight rungs of the ladder
Big-O classifies the asymptotic growth of a function, not the wall-clock time. The ladder Bugdle uses has eight rungs ordered from cheapest to most ruinous: O(1), O(log n), O(n), O(n log n), O(n²), O(n³), O(2ⁿ), O(n!). The puzzle's answer sits at rung 8 of 8. The four-guess budget plus higher/lower hints means the puzzle is solvable in at most ⌈log₂(8)⌉ = 3 perfectly-read guesses; the fourth attempt absorbs misreads of the snippet.
Common confusables
Nested loops aren't automatically quadratic — the inner loop has to range over n, not a constant. A loop that always runs ten iterations is still O(1) work per outer iteration. Similarly, a recursive function isn't automatically exponential just because it calls itself twice — if the recursion has overlapping subproblems and you memoise, you collapse back to polynomial. Always ask: what does n really mean here, and how many distinct subproblems are there?
External reference: Big O notation — Wikipedia.