Are there infinite possible values of $1^i$ where $i=\sqrt{-1}$ Are there infinite possible values of $$z=1^i$$

Since $$e^{i2k\pi}=1$$ we have

$$1^i=\left(e^{i2k\pi}\right)^i=e^{-2k\pi}$$

so for different values of $k \in \mathbb{Z}$ we get different values. Is this concept correct?
You cannot use power laws of the reals for complex values. They do not hold on that wider space. So $$\left(e^{i2k\pi}\right)^i\not=e^{-2k\pi}$$

Otherwise you can easy "proof" that $-1=1$ by $$-1 = e^{-i\pi} = \left(e^{-2i\pi}\right)^\frac{1}{2} = 1^\frac{1}{2} = 1$$

So you assume a false condition…
I agree with
the OP's statement that
$1^i
=\left(e^{i2k\pi}\right)^i
=e^{-2k\pi}$
for any integer $k$.
For $k=0$,
$1$ is one of the possible values.

To "check",
note that
$(1^i)^{-i}
=(1)^{i\cdot (-i)}
=1^1 = 1
$
and
$(e^{-2k\pi})^{-i}
=e^{(-2k\pi)(-i)}
=e^{2k\pi i}
=1
$.