= Solution
Use the <principal complex logarithm> at each stage of the nested <complex exponentiation>. Since $\operatorname{Log}i=i\pi/2$,
$$
i^i=e^{i\operatorname{Log}i}=e^{-\pi/2}.
$$
This inner value is positive real, so its <principal complex logarithm> is $-\pi/2$. Consequently
$$
\boxed{(i^i)^i=e^{i(-\pi/2)}=-i=0-i.}
$$
One must retain the parentheses: <complex exponentiation is not associative>. If all <complex logarithm> branches are allowed, the inner values are $e^{-\pi/2-2\pi k}$. Taking the outer logarithm with an independent integer branch index $\ell$ yields $\exp[-i\pi/2-2\pi ik-2\pi\ell]=-ie^{-2\pi\ell}$. Thus the complete multivalued set is $\{-ie^{-2\pi\ell}:\ell\in\mathbb Z\}$, and the boxed answer selects its principal evaluation.
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