Finitely iterated prisoner's dilemma is just like the traveler's dilemma, on which see this article by Kaushik Basu. The "always defect" choice is always a (in fact, the only) Nash equilibrium and an evolutionarily stable strategy, but it turns out that if you measure how stable it is, it becomes less stable as the number of iterations increases. So if there's some kind of noise or uncertainty (as Dagon points out), cooperation becomes rational.
Above link is bad, try this.