Does every perfect information game in extensive form have a Nash equilibrium?

Question

I have been struggling to reason whether this is true. Given a perfect information in extensive form (say chess or many such board games), is it guaranteed that there exists a pure strategy Nash equilibrium? If not, what would be a counter example?

Answer 1

Every finite game with perfect information in extensive form has a pure nash equilibrium.

A standard proof would be based on backward induction: We go to the final nodes of the game (i.e., the ones where a player acts and all his actions terminate the game), and there in equilibrium, the player will choose the action that is best for him (otherwise, he has a profitable deviation). Then we replace each such decision node with the terminal payoff. We obtain a smaller tree. By repeating this process, we obtain an equilibrium in pure strategies.

Another option is to prove it by induction on the number of nodes in the game (for 1 it is trivial, if it is true for any game of length less than $n$, do it for each subgame starting after the root, then replace each such subgame with the equilibrium payoff and concatenate the strategies).