algebra precalculus - Why does this work for $ i^{2i} $?

I'm finding the principal value of $$ i^{2i} $$

And I know it's solved like this:

$$ (e^{ i\pi /2})^{2i} $$

$$ e^{ i^{2} \pi} $$

$$ e^{- \pi} $$

I understand the process but I don't understand for example where does the $ i $ in $ 2i $ go?

Is this some kind of a property of Euler's number? if so please explain to me.

Answer

$$\bigl(e^{i\pi /2}\bigr)^{2i} = e^{(i\pi /2) \cdot 2i} = e^{i^2\pi}.$$

This is just an application of the exponent laws. Don't overthink it!