I was watching Lennart Augustsson's MicroHs, a tiny Haskell Compiler talk, and saw that he mentioned a pretty neat equality: $$log(i^i) + \frac{\pi}{2} = 0$$, and punctued it with a casual "it's obvious if you think about it." And even though I dislike trigonometry and complex numbers, it's indeed pretty trivial, thanks to Euler's identity, as usual when dealing with exponentials and complex numbers:
%% \begin{aligned} e^{i\pi} = -1 &\iff e^{i\pi} = i^2\\ &\iff \log{e^{i\pi}} = \log{i^2}\\ &\iff i\pi = 2 \log{i}\\ &\iff \pi = 2\frac{\log{i}}{i}\\ &\iff \frac{\pi}{2} = \frac{\log{i}}{i}\\ &\iff \frac{\pi}{2} = -\log{i^i}\\ &\iff log(i^i) + \frac{\pi}{2} = 0\\ ■ \end{aligned} %%
Nothing too complicated indeed, and you can use this to get the value of $$i^i$$, which is $$e^{-\frac{\pi}{2}}$$.