Why does $i^2 = -1$? I understand that $i = \sqrt{-1}$, but it seems as if $i^2$ would be: 

$$\sqrt{-1}\sqrt{-1} = \sqrt{1} = \pm 1$$

Why is this not the case?
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.
if you square the square root of 'something', the result is equal to the 'something'
i.e. the (square root of two) squared is equal to two