I saw the following property in wikipedia \begin{align} (e^z)^n=e^{zn} \tag{1} \end{align} where $z \in \mathbb{C}$ and $n \in \mathbb{Z}$. It's possible to change the $e$ in $(1)$ to some other number ? For example:
\begin{align} (a^z)^n=a^{zn} \end{align}
where $z \in \mathbb{C}$, $n \in \mathbb{Z}$ and $a$ is maybe a real number or even complex? Or even change the conditions on $z$ and $n$.
Another thing related to $(1)$. Gelfond's theorem says that $a^b$ is transcendental if $a$ is algebraic and $a \neq 1$, $a \neq 0$ and $b$ is irrational. For example $2^{\sqrt 2}$ is transcendental. One result from this theorem is that $e^\pi$ (Gelfond constant) is transcendental. Noting that:
\begin{align} & e^{i\pi} = -1 \\ & (e^{i\pi})^{-i} = (-1)^{-i}\\ & e^{\pi} = (-1)^{-i} \end{align}
since $(-1)$ is algebraic and $-i$ is irrational the theorem applies. But in the second line above didn't we just violate the property $(1)$ because $-i$ in not an integer?
Given two complex numbers $a,z\in\mathbb{C}$, we have $a^{z}=e^{z\log a}$. You are correct that for any $a,z\in\mathbb{C}$ and $n\in\mathbb{Z}$, $\left(a^{z}\right)^{n}=a^{zn}$, which can be verified by substituting $e^{z\log a}$. If $n$ is no longer an integer, then it is no longer true because it becomes multivalued.
Using your example, we have \begin{align*} \left(-1\right)^{-i}=e^{-i\log\left(-1\right)}=e^{-i\left(\ln\left|-1\right|+i\arg\left(-1\right)\right)}=e^{-i\left(0+i\arg\left(-1\right)\right)}=e^{-i^{2}\left(\pi+2\pi k\right)}=e^{\pi+2\pi k},\,k\in\mathbb{Z}. \end{align*}