OK, I agree that this is the standard solution. Here's my interpretation:
Start counting days d on the day the Guru speaks. Each islander knows that if the observable blue eyed islanders n are greater than d then that islander must leave at midnight. On day 1 the blue eyed islanders n value is one less than the n' for those without blue eyes. When n' equals d then the blue eyed islanders will leave.
n' = n + 1
Leave IFF d > n (supposing no one else has left yet)
If n on d_1 was more than it is on d_2 then islander knows they don't have blue eyes and won't leave because other islanders cannot leave with certainty when d < or = n. Again, leave if and only if d > n but since n' is strictly greater than n ... those w/ non-blue eyes won't leave that day which happens to be when d = n' ... And if anyone left the night before (for d>1, to be clear) then the islander may conclude that they do not have blue eyes. So, on d=100, the blue eyed islanders each see n=99 each reasoning that "if I didn't have blue eyes the rest would've left yesterday". And when any leave, the remaining islanders each reason that "if they were certain, they must have observed fewer other blue eyed islanders than days counted ... I did not. So I must not have blue eyes." The floor of (n<d) is the floor of (n'-1 <d). Islanders leave as soon as they are certain and that's the floor of the function only. If any leave, they know they're operating with different values and therefore have insufficient data to ever conclude the color of their eyes; since they know the Guru will never again speak.
As mentioned by /u/Mendoza2909, knowing with certainty that your eyes aren't one color doesn't imply with necessity that they are any other specific color. Which is necessary to know in order to leave, apparently.
However, what happens on day 99 that is any different than the day before? So why don't the 100 blue eyed islanders leave earlier? There is a presumed coordination between them that is based on counterfactual cases that are empirically untrue. Does logic really get anyone off the island?
Day 1: Islander's Log
"The Guru has spoken! At long last, we have their information and might finally leave this wretched place of silence. I have heard tell of magical objects, called mirrors, off in distant lands. They are shiny beyond reason and when a person gazes upon it, they see the color of their own eyes. I wish to see such a wonder of science before I die."
"What follows is my reasoning recorded at once, to preserve for prosperity this blessed occasion; for I am miffed. The Guru has stated that she 'sees someone with blue eyes'. Except, that is nothing new. I have seen 99 people with blue eyes since I first counted them. Indeed, if there were but 1 islander before her with blue eyes, she could have said the same. We were all there. And if there were 1 of blue eyes, they would be leaving this very night. Indeed, if there were 2 they could leave tomorrow with this new, sacred information."
"This is counterfactual induction, OK. I see 99. Does that mean I should wait until day 100? It is just the case that each of the blue eyes sees 98 or 99 others. If I am blue eyed, they see 99. Meaning we leave the day after our own count, if no one else has left yet. For once islanders leave, the count will have been certain and the rest left waiting must have seen every blue eyed islander for the last time."
"There are brown eyes and green eyes visible to every islander except the Guru. She only sees brown eyes left. Could she conclude that her eyes are brown on day 101? Wait, I almost forgot about that poster of 'The Many Eye Colors of the World' hanging in the visitors center. Of course, she must suppose the possibility that she has green or brown or red or any colored eyes ... until she has sufficient information to the contrary. Also, it's a good thing that we all know that we each have the same color vision, such that the Guru's categorization of 'blue' is the same as anyone else's on the island. If a blue eyed islander was blue-colorblind, this whole business would be over at that."
"Now, the philosophical problem is ensuring coordination. How do I know they have each reasoned in the same manner? Hypothetically, new information is added with each passing day. Except that from day 1 thru 98, I know the same and certainly so does everyone else. If no one leaves tonight, it would not necessarily be because the Guru was referring to me ..."
"We all have the same info from day 1. There are too many blue eyes. We gain nothing on day 2. It would be overusing counterfactual reasoning to say that we do. The chain is broken. We cannot use untrue information to inductively assert a certain conclusion, can we?! IS THIS REASONING FALLACY?! HAS THE GURU TOLD US NOTHING?!!"
"Consider the maximum number of blue eyed islanders there could have been, given her statement, and still been able to leave at some point. If 1, then they could leave on the first day. The Guru does not lie. That individual would not see any blue eyes, and conclude they have blue eyes. If 2, then each would see one other, and suppose on that first day that other would follow the first case or else both had blue eyes. If 3, they would all see two others on day one ... they would all take on the reasoning of the second case. Thinking the others would leave on day 2 if not blue-eyed themselves, all three would leave on day 3. OK, now 4. Can this reasoning continue?"
"Where is the flaw? On day one, 4 blue eyed islanders see 3 each and suppose the reasoning of the third case until day 4 ... the three others, they reason with certainty, see exactly 2 or 3 others ... and so, are presumably using the third case also, to wait for more information on day 4, to leave if no one left, which eventually would be the case, wouldn't it?"
"In the case of 4 blue eyed islanders, no one is reasoning that anyone is reasoning that someone is possibly only seeing one set of blue eyes. They all know that no one is worried about that first day dooming them to stay on this stupid island forever. No one suspects anyone might leave on the first day. What stops them from being pragmatic in their counterfactual induction? OK, cool. While it may be nice to try and have a simply coordinated date of exit, we want to get out of here sooner than later. Let's coordinate on that instead of some stupid three month wait."
"How do we do that though? What if ... We reduce the days to the lowest number of blue eyes possibly seen. I see 100. If I have blue eyes, that's 101 islanders. No one would see less than 100. Reduce 100 to the case of 1 blue eyed islander and 101 to the case of 2. Those who see 101 will reduce their count to day 1 and 102 to day 2. None of us have the information to discern any difference between. None are certain on day one, leading to faulty reasoning on day 2. We would all leave whether we had blue eyes or not."
"That would be great! It would be what we really want. But why wait a day? If all we really want is to leave, then let's just all leave tonight. ... except that ... Except that we want certainty. We want to know for sure and this stupid puzzle is the only way to find out. We can count ferries. We can count eyes. We can reliably coordinate based on that information. We'll all wait to leave if no one else has as soon as the visible count is less than the number of days. Another way to put it is to leave if the current count of blue eyes is less than the number of days. Let's be certain. I see 100, so I will leave with certainty on day 101 if no one has left already."
Day 101
"Fuck."