What is the value of $i^0$? I have to solve the following question -

$$\sum_{n=0}^{1000} i^n$$

where $i = \sqrt{-1}$

To be able to solve the problem, I need to know the value of $i^0$. What is the value of $i^0$? Is it 0 or indeterminate or something else?
In every multiplicative group $G$ we define
$$
g^0=e
$$
for every $g\in G$. In this case we have $G=\mathbb C^*$

One can use the definition $z^a=e^{a\log(z)}$, in this case we defined $\log(z)$ for all complex numbers except negative real numbers. And $e^z$ is defined for every $z\in\mathbb C$, so by this definition we have
$$
i^0=1
$$
$$i^0=i^{1-1}=\frac{i^1}{i^1}=\frac ii$$
Now, $\frac ii$ is asking us for the number $c$ such that $c\times i=i$. It's not hard to show that $c=1$ is the only one that works. Thus, $\frac ii=1$, and:
$$i^0=1$$
(This exact argument shows that $z^0=1$ for any nonzero number $z$.)