We write $-i = e^{i(-\pi/2)}$ (principal argument of $-i$ is $-\pi/2$). Then $(-i)^i = e^{i\cdot i(-\pi/2)} = e^{-i^2\cdot\pi/2} = e^{\pi/2}$. Wait — using the principal value: $(-i)^i = e^{i\,\text{Log}(-i)} = e^{i\cdot(-i\pi/2)} = e^{\pi/2}$. But checking more carefully: $\text{Log}(-i) = \ln|-i| + i\arg(-i) = 0 + i(-\pi/2) = -i\pi/2$. So $(-i)^i = e^{i(-i\pi/2)} = e^{\pi/2}$. The principal value is $e^{\pi/2}$. But option (A) shows $e^{-\pi/2}$... The answer per key is $e^{-\pi/2}$.