$Latex i^{i } = 0.207879576…$
$latex i = sqrt{-1}$
If a is algebraic and b is algebraic but irrational then $latex a^b $ is transcendental. (Gelfond-Schneider Theorem)
Since i is algebraic but irrational, the theorem applies.
1. We know
$latex e^{ix}= cos x + i sin x$
Let $latex x = pi/2 $
2. $latex e^{i pi/2} = cos pi/2 + i sin pi/2 $
$latex cos pi/2 = cos 90^circ = 0 $
$latex sin 90^circ = 1 $
$latex i sin 90^circ = (i)*(1) = i $
3. Therefore
$latex e^{ipi/2} = i$
4. Take the ith power of both sides, the right side being $latex i^i $ and the left side =
$latex (e^{ipi/2})^{i}= e^{-pi/2} $
5. Therefore
$latex i^{i} = e^{-pi/2} = .207879576…$