Euler's Formula:

e^(ix) = cos(x) + i*sin(x)

If x is Pi/2 then:

e^( i * Pi / 2 ) = cos( Pi/2 ) + i * sin ( Pi/2 )

e^( i * Pi / 2 ) = 0 + i

e^( i * Pi / 2 ) = i

raise both sides to the power i like this:

( e^( i * Pi / 2 ) ^ i = i ^ i

simplify

e^( i * i * Pi / 2 ) = i ^ i

e^( -1 * Pi / 2 ) = i ^ i

e^( -Pi/2 ) = about 0.20788, so you could say i ^ i = about 0.20788, but that's not all

the reason Pi/2 worked is because cos(Pi/2) = 0 and sin(Pi/2) = 1

we can go around a full circle and see that cos(5Pi/2) = 0 and sin(5Pi/2) = 1

9Pi/2 would also work, 13Pi/2 would work, and so on

the form ends up being (4n+1)Pi/2, where n is any integer (works for negative and positive integers)

Answer: e^( -(4n+1)Pi/2 ) where n is any integer

Extra reading: http://www.math.hmc.edu/funfacts/ffiles/20013.3.shtml