Prove: In his winning strategy, player 1 (P1) in an $n \times n$ Chomp game must choose $(2,2)$
(See below for chomp description)
In an $n \times n$ Chomp game, we know that by choosing $(2,2)$, P1 secures his winning, because afterwards he could simply make symmetric moves to P2.
I'm trying to prove that $(2,2)$ must be the first move for P1 in order for him to win.
I've tried to proof by contradiction, but couldn't solve this. I've also tried by induction, but couldn't reduce the $n \times n$ board to a smaller squared board.
The Chomp game description, as was given here:
The game of Chomp is played by two players. In this game, cookies are laid out on a rectangular grid. The cookie in the top left position is poisoned. The two players take turns making moves; at each move, a player is required to eat a remaining cookie, together with all cookies to the right and/or below (that is all the remaining cookies in the rectangle, in which the first cookie eaten is the top left corner). The loser is the player who has no choice but to eat the poisoned cookie. Prove that if the board is square (and bigger than 1 × 1) then the first player has a winning strategy.