scott wrote:
>> So the answer we seek is (approximately) 2↑↑↑2.> > Isn't that 4?
No, since 2^^^2 = 2^^2^^2 = 2^^(2^2^2) = 2^^16 =
2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2.
However, in this case you can very easily give a somewhat tighter lower
bound of 2^^2048, that is 2^2^2^2^ ... 2048 times ... ^2^2^2
