complex numbers - Are both square roots of -1 valid in Euler's Identity?

In every case I've ever seen, Euler's Identity is written as

$e^{i\pi} + 1 = 0$

with the "positive" $\sqrt{-1}$. However, my understanding is that both $i$ and $-i$ are valid for $\sqrt{-1}$.

Does this mean that

$e^{-i\pi} + 1 = 0$

is also a valid identity?

Answer

Yes.

$$e^{-i\pi} =\cos (-\pi)+i\sin(-\pi)=-1$$