The cursed chocolate

1. Don't look for the exact winning play: first decide if the first player theoretically wins or loses.
2. Imagine if the second had a winning strategy and see if the first could steal it.
3. The test demonstrates the existence of strategy, not a practical recipe to always play.

Answer: the first player has a winning strategy. The demonstration does not always give us the specific move that the first player must make, but it does prove that a winning move exists. First we observe that the game is finite: in each turn at least one square disappears. Also, there are no ties. Therefore, from any position, either the player whose turn it is can force victory, or he cannot. Now the first player considers a minimum move: eating only the square in the lower right corner. That play is legal and does not touch the poisoned square. There are two cases. Case 1: after that move, the second player is in a losing position. Then the first player wins starting like this. Case 2: after that move, the second player is in a winning position. Then the second has some response that leaves the first in a losing position. But the second's response consists of choosing a square and eating everything that is below and to the right. That same choice was also legal from the initial board. Also, any such bite already includes the bottom right corner, so directly doing that bite from the beginning produces the same result as: 1. eat first only the lower right corner;

1. then make the winning answer of the second. Therefore, if the second player had a winning response after the minimum move, the first player could have made that same bite as the first move. So, in either case, the first player has a winning move. On a $2\times 2$ board, for example, removing the bottom right corner does win outright. But on larger boards it is not necessary for this to always be the winning move: if it is not, the previous argument shows that there is another first winning move.