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!