> Ok, so let me ask you: How does your picture show that e^pi > pi^e?
I don't know what Joe's explanation is (when more hand-holding details are included), although I haven't tried very hard to figure it out. However, one way is to note that the numerical order of e^pi and pi^e is unchanged if e^pi is replaced with (ln e)/e and pi^e is replaced with (ln pi)/pi. This follows from the fact that the natural logarithm function is strictly increasing:
e^pi # pi^e [# is < or = or >, whichever is correct]
Off hand, I don't see how we can tell by precalculus methods which of 1/e and (ln pi)/pi is greater (or equivalently, which of pi/e and (ln pi) is greater), but I do know that taking the derivative of f(x) = (ln x)/x and setting the result equal to zero (and paying attention to where the derivative is positive and where the derivative is negative), we find that f(x) has a maximum (local and global) at x = e. Thus,
(ln e)/e > (ln pi)/pi
and so we get
e^pi > pi^e
See the following web page for several proofs of this inequality, including a usefully labeled graph of f(x) = (ln x)/x.