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Hello, Mhmh96!
$\displaystyle \text{Prove: }\:i^i \:=\:\dfrac{1}{\sqrt{e^{\pi}}}$
$\displaystyle \text{Let }\,z \:=\:i^i$
$\displaystyle \text{Take logs: }\:\ln(z) \:=\:\ln(i^i) \quad\Rightarrow\quad \ln(z)\:=\:i \ln(i) \;\;{\bf[1]}$
$\displaystyle \text{Given: }\:e^{i\theta} \:=\:\cos\theta + i\sin\theta$
$\displaystyle \text{Let }\theta = \tfrac{\pi}{2}\!:\;\;e^{i\frac{\pi}{2}} \;=\;\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} \quad\Rightarrow\quad e^{i\frac{\pi}{2}} \:=\:i $
$\displaystyle \text{Take logs: }\:\ln\left(e^{i\frac{\pi}{2}}\right) \:=\:\ln(i) \quad\Rightarrow\quad i\tfrac{\pi}{2}\ln(e) \:=\:\ln(i) $
. . $\displaystyle \text{Hence: }\:\ln(i) \:=\:i\tfrac{\pi}{2}$
$\displaystyle \text{Substitute into }{\bf[1]}\!:\;\;\ln(z) \:=\:i\left(i\tfrac{\pi}{2}\right) \quad\Rightarrow\quad \ln(z)\:=\: -\tfrac{\pi}{2}$
$\displaystyle \text{Therefore: }\:z \;=\;e^{-\frac{\pi}{2}} \;=\;\dfrac{1}{e^{\frac{\pi}{2}}} \;=\; \dfrac{1}{(e^{\pi})^{\frac{1}{2}}} \;=\;\dfrac{1}{\sqrt{e^{\pi}}} $