This article is from the Puzzles FAQ, by Chris Cole firstname.lastname@example.org and Matthew Daly email@example.com with numerous contributions by others.
R = N ^ (N ^ (N ^ ...)). What is the maximum N>0 that will yield a finite R?
Let N be the number in question and R the result of the process. Then
R can be defined recursively by the equation:
(1) R = N^R
Taking the logarithm of both sides of (1):
(2) ln(R) = R * ln(N)
Dividing (2) by R and rearranging:
(3) ln(N) = ln(R) / R
(4) N = R^(1/R)
We wish to find the maximum value of N with respect to R. Find the
derivative of N with respect to R and set it equal to zero:
(5) d(N)/d(R) = (1 - ln(R)) / R^2 = 0
For finite values of R, (5) is satisfied by R = e. This is a maximum of
N if the second derivative of N at R = e is less than zero.
(6) d2(N)/d2(R) | R=e = (2 * ln(R) - 3) / R^3 | R=e = -1 / e^3 < 0
The solution therefore is (4) at R = e:
(7) Nmax = e^(1/e)