Why does $i^2 = -1$?

If I understand correctly, you're thinking is as follows: $$\ \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = \pm 1. $$

I'll address the primary question here: in writing this, you are assuming the "rule" that tells us: $$ \sqrt{ab} = \sqrt{a}\sqrt{b} $$ also applies to $a, b$ when both are negative. However:

$\sqrt{ab} = \sqrt a\sqrt b\;$ if and only if $a$ and $b$ are not both negative real numbers.

The usual laws of exponents (and roots) do not apply to complex imaginary numbers, nor, in the case of the square root, do they apply to negative numbers of any sort.

The rule that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only holds when $a, b$ are positive.

To add some more context: these rules hold only for positive numbers for a few reasons, but in general, the existence of complex numbers is something necessitated by an inadequacy of the reals to solve certain types of problems.

In a way, we might look at $i^2 = -1$ as a definitive property of $i$ (that is, a property that holds because we define it to be so), rather than a consequential property.

Complex arithmetic is essentially synthetic: it's something we made up. It happens that the rules we chose are very useful and have exceptionally nice properties, so we stick with them.

But in the end, we define $(a,b)(c,d) = (ac-bd,ad+bc)$. This is exactly equivalent to defining $i^2 = (0,1)(0,1) = (-1,0)$.

A main misconception here is that $\sqrt{a^2}=\pm a$. That is not true, but $\sqrt{a^2}=|a|$ is true for any real number $a$.

What you are confusing it with is solving the equation $x^2=c$. Some students are taught to "take the square root of both sides, but to avoid dropping a solution, use $\pm$":

$$x=\pm\sqrt{c}$$

It's important not to confuse these two things: finding the square root of a number, and solving an equation of the form $x^2=c$.