The chocolate bar is a rectangle $m*n$ , divided by a recess into single squares. Two players play the next game in turn. On each turn, it is allowed to pick up one square of chocolate that has not yet been taken, as well as squares that have not yet been taken, lying on the right and above (including squares, lying above in the same column and on the right in the same row). The one who occupies the last square loses. For all natural $m$ and $n$, specify which player has the winning strategy.
I`m trying to solve it by contradiction, assuming that the second player has a winning strategy. (The first player takes the top rightmost square and then uses second player strategy). But it seems like the wrong way.
Should I reduce it to reverse Nim game? Or can it be solved in a simpler way?