**Hint:** Think Euler...or De Moivre...or...

**Solution Commentary:** By Euler's equation, e^{ix} = cos(x) + i sin(x).

For x = pi/2, we get e^{(i*pi)/2} = cos(pi/2) + i sin(pi/2) = 0 + i = i.

Now, raising each side to the ith power, e^{(i*i*pi)/2} = i^{i}.

But, i*i = -1, which implies that e^{(-pi)/2} = i^{i} is a real number.

Finally, e^{(pi)/2} > e^{(-pi)/2} where e^{(pi)/2} = i^{-i}. Thus, i^{-i} > i^{i}.

For a variation on this argument, see Churchill's Complex Variables and Applications (1960), pp. 60-62.