0.207879576...

iⁱ = 0.207879576... (i^î)  (Here i is the imaginary number sqrt(-1). Isn't this a real beauty? How many people have actually considered rasing i to the i power? If a is algebraic and b is algebraic but irrational then a^b (a^b) is transcendental. Since i is algebraic but irrational, the theorem applies. Note also: iⁱ is equal to e^{(- pi / 2 )} and several other values. Consider iⁱ = e^{(i log i )} = e^{( i times i pi / 2 )} . Since log is multivalued, there are other possible values for iⁱ.
Here is how you can compute the value of iⁱ = 0.207879576...

1. Since e^(ix) = Cos x + i Sin x, then let x = Pi/2.
2. Then e^(iPi/2) = i = Cos Pi/2 + i Sin Pi/2; since Cos Pi/2 = Cos 90 deg. = 0. But Sin 90 = 1 and i Sin 90 deg. = (i)*(1) = i.
3. Therefore e^(iPi/2) = i.
4. Take the ith power of both sides, the right side being i^i and the left side = [e^(iPi/2)]^i = e^(-Pi/2).
5. Therefore i^i = e^(-Pi/2) = .207879576...