We know that the following conditions are true:
$(-1)^{2n}$ is $1$ where $n$ is an integer
$(-1)^{2n+1}$ is $-1$ where $n$ is an integer.
We can extend this reasoning for rational numbers.
If we let a number be written in the form of $a+\frac{b}{c}$ such that $\gcd(b,c) = 1$ and $a, b, c$ are integers, then to evaluate $(-1)^{a+\frac{b}{c}}$, we can separate $(-1)^{a+\frac{b}{c}}$ into $(-1)^{a}\times\frac{(-1)^b}{(-1)^c}$. The terms $(-1)^{a}$, $(-1)^b$, $(-1)^c$ can be reduced to either of the two cases above; and we solve for $-1$ or $1$.
Now what is $(-1)^\pi$. It isn't rational, so there isn't a perfect ratio that can be done with the method above. I'm suspecting that roots of unity come to play here or DeMoirve's Theorem, but don't they resolve to whether or not you can solve $r^n$ where r is the radius of the complex number and n is exponent.
NOTE: I'm still in high school, so I am looking for an answer that does not have any advance math- but will gladly accept any response to the question for curiosity's sake.
Thank you :)
For any complex numbers $z$ and $a$, the term $z^a$ is defined as $z^a=e^{a\log(z)}$. Then, with $z=-1$ and $a=\pi$ we have
$$\begin{align} (-1)^\pi&=e^{\pi\log(-1)} \tag 1\\\\ &=e^{\pi(i\pi +i2n\pi)} \tag 2\\\\ &=e^{i\pi^2 (2n+1)}\\\\ &=\cos(\pi^2 (2n+1))+i\sin(\pi^2 (2n+1)) \end{align}$$
for all integer values of $n$.
Note that in going from $(1)$ to $(2)$ we recognized the multivalued nature of the complex logarithm, $\log(z)=\log(|z|)+i(\text{Arg}(z)+2n\pi)$. Here, $|z|=1$ and $\text{Arg}(z)=\pi$.