Blue Eyes - "The Hardest Logic Puzzle in the World"

[lilphil1989]

---

I think that 100 people kill themselves on the 100th night.

Here's my reasoning:

(1) Start with a situation in which there's one blue-eyed person.
This person looks around, sees no other blue-eyed people and can therefore conclude that, since there is at least one blue-eyed person, there is exactly one: himself. He then kills himself on the first night.

(2) Now imagine a situation in which there are 2 blue-eyes. Each blue-eye will look around and see one other blue-eye. He can then follow the reasoning of (1) and conclude that if he himself has brown eyes, the other blue-eye that he sees will kill himself on the first night. When the other blue-eye doesn't kill himself on the first night (because each can see another blue-eyed person), both blue-eyes people will realise that there is one more blue-eyed person: himself. Therefore they both kill themselves on the second night.

(3) 3 blue-eyed people. Each follows the reasoning of (2), and when the other blue-eyed people don't kill themselves on the second night, all 3 kill themselves on the third.

(4)In general, for the case of N blue-eyed people, they all refer back to the reasoning of the (N-1)th case and will all kill themselves on the Nth night.

---

Is this correct?

Given the title of the post, I think I may have missed some subtlety :S
Or maybe something really obvious!