I misread this question and began thinking about the value $e^\pi$. This lead me to the Wikipedia on Gelfond's Constant, which suggests deriving a numerical value for $e^{\pi}$ by using Euler's identity: $$e^{i\pi}=-1\\ \sqrt[i]{e^{i\pi}}=\sqrt[i]{-1}\\ e^{\pi}=(-1)^{-i}\approx 23.14069$$
My question is, how can we go about taking an $i$th root? This just doesn't seem like something that would be well-defined. This congruence along with the numerical approximation is mind-boggling.
In general, if $y,z$ are complex numbers, we don't have a single value of $y^z$. Rather, we have to take a value of $\log y$, and compute $e^{z\log y}$. Unfortunately, there can be infinitely many values of $\log y$. If $e^w=y$ then $e^{w+i2\pi k} = y$ for all integers $k$.
Thus we can write $\log -1 = i(\pi + 2\pi k)$ and thus $(-1)^i = e^{-\pi-2\pi k}$. So the possible values for $(-1)^i$ are of the form $e^{\pi n}$ where $n$ can be any odd integer.
You can see this as an extension of the idea that there is more than one square root of $2$. In general, there are infinitely many possible values for $y^z$, but if $z$ is an integer, the different values of $\log y$ yield the same value, and if $z$ is a rational number, $y^z$ can only take finitely many values. However, the general case there are infinite possible values of $y^z$.