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In page Prisoner's dilemma:

"

If the iterated prisoner's dilemma is played a finite number of times and both players know this, then the dominant strategy and Nash equilibrium is to defect in all rounds. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on. The same applies if the game length is unknown but has a known upper limit.[citation needed]