On a hot summer day, n equally bright philosophers, tired from all that philosophising, were napping in an orchard. A prankster came by, and painted all of their faces black with charcoal.

When the philosophers woke up, they started laughing at the others... until they suddenly realised all of their faces must be black!

How did they come to that conclusion?

By analyzing this problem at various known values n, the unknown variable N problem can be solved.

For 2 persons, there n is 2.  Person k=n is the analyzer.  Person k looks at each person k=1 to k=n-1 to see what they are doing.  Person k=1 is laughing.  He can't be laughing at himself, but must be laughing at a person in the remaining group.  Assuming himself to be blemish free, person k=n realizes that there are n-2 people that person k=1 could be laughing at.  In this case, n=2, so n-2=2-2=0 people, that person k=1 could be laughing at, if person k=n were blemish free.  Since he is laughing at someone, then the assumption that person k=n is blemish free is obviously false.

For 3 persons, there n is 3.  Again, person k=n is the analyzer.  Person k looks at each person, k=1 to k=n-1, to see what they are doing.  Person k=1 is laughing.  He can't be laughing at himself, but must be laughing at a person in the remaining group.  Assuming himself to be blemish free, person k=n realizes that there are n-2 people that person k=1 could be laughing at.  In this case, n=3, so n-2=3-2=1 person, that person k=1 could be laughing at, and that person must be k=2, as k=3 believes himself to be blemish free and person k=1 can't laugh at himself.  Now, analyzing k=2, person k=n realizes that again there are n-2 people that can't be laughed at, in this case, them being k=n and k=2.  That only leaves k=1 as the person that k=2 is laughing at.  This leads to a loop, until person k=n realizes that if k=1 is laughing, then he too believes himself to be blemish free, as does person k=2, since he also is laughing.  This means that there are n-3 people that cannot be laughed at.  With n=3, that leaves us with n-3=3-3=0 people that can be laughed at, if all people involved believe themselves to be blemish free.  Since they are laughing at someone, then at least one of them must be laughing at person k=n, so the assumption that person k=n is blemish free is false.

As this continues, a pattern develops, where there are n people who believe they are blemish free, with n people laughing, leaving n-n=0 people to be laughed at, therefore, all must be blemished.

To put it more mathematically:

For each person k of group N, there exists a person k-sub-1 such that person k-sub-1 is also in group N, leaving N-2 people to be laughed at by either k or k-sub-1.  This progresses with N approaching 1 such that the last group consists of no other person at which to laugh.  At this time, that person ceases to laugh, because he realizes he himself is the only one at which there is to laugh at, and therefore, he must be through some permutation or another, at one point eliminated as a person who was laughed at, and therefore he has himself been duped by the prankster.