The key of the problem includes the statement that all the islanders have seen all the other islanders and know their eye colors. From there we have the simple

The key of the problem includes the statement that all the islanders have seen all the other islanders and know their eye colors. From there we have the simple case of n = 1: suppose Bob is the only blue-eyed islander. He's seen everyone else's eyes, knows nobody else has blue eyes, and so can conclude from n >= 1 that in fact n = 1 and he is that one. For the case of n = 2: suppose Alice and Bob are the two blue-eyed islanders. Alice knows Bob is blue-eyed but doesn't know she is. Bob knows Alice is blue-eyed but doesn't know he is. There are two possibilities: n = 1 or n = 2, since they have seen every other person's eyes and know there are no other blue-eyed people. Each can watch the other's behavior on the first day. But since neither one knows their own eye color yet, neither one commits suicide that day. Then each one realizes that for the other not to commit suicide, it must be the case that n > 1. Since they know every other person's eye color, the only possible explanation is "n = 2 and I'm the other one". They commit suicide on day 2. From there we can just apply the same reasoning every time. If n = 3, Alice, Bob and Carol each look for the other two to commit suicide on the second day (from the reasoning for n = 2). When they don't, the only possible conclusion is "n = 3 and I'm the third one" and they all commit suicide on day 3. That line of reasoning then works with any successive value for n: if there are n blue-eyed islanders, each looks for the other n - 1 islanders to commit suicide on day n - 1. When that doesn't occur, the only possible conclusion those people can reach is "there's one more blue-eyed islander, and it's me". Then all n of them commit suicide on day n.