\begin{align}
i&= \text{cos}\left(\frac{4n\pi+1}{2}\right)+i\text{sin}\left(\frac{4n\pi+1}{2}\right)\ (n∈\mathbb{Z})\\
i^i&=\text{cos}\left(\frac{4n\pi+1}{2}i\right)+i\text{sin}\left(\frac{4n\pi+1}{2}i\right)\\
&また，e^{i\theta}=\text{cos}\theta+i\text{sin}\theta より\\
i^i&=e^{i\left(\frac{4n\pi+1}{2}i\right)}\\
&=e^\left(-\frac{4n\pi+1}{2}\right)\\
&=\frac{1}{e^{\frac{4n\pi+1}{2}}}∈\mathbb{R}
\end{align}
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