In the solution to the problem below, why does B win if A takes l stones, $1\leq l < 2^k$ (it's in the proof of sufficiency)? Doesn't that make the situation $(a, a)$ where $a > 2^k$? And if so, would the inductive hypothesis be insufficient to conclude who will win?
Consider the "subgame" consisting of the first $2^k$ stones off the heap. We suppose A takes $l$ stones with $1 \leq l \leq 2^k$. We're now at position $(2^k, l)$. By our inductive hypothesis, B wins. Having removed $2^k$ stones in total from the heap, we're now at the game with starting position $(2^k, m)$, where $m \leq l$. Again by our inductive hypothesis, B wins this game. So B wins the entire game.