Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$? We know $$i^2=-1 $$then why does this happen?
$$
i^2 = \sqrt{-1}\times\sqrt{-1}
$$
$$
=\sqrt{-1\times-1}
$$
$$
=\sqrt{1}
$$
$$
= 1
$$
EDIT: I see this has been dealt with before but at least with this answer I'm not making the fundamental mistake of assuming an incorrect definition of $i^2$.

There are many ways to show that your second equality is incorrect. Just for the sake of argument and a refreshing change of pace, suppose your point is true. That is, suppose $i^{2} = 1$. Then
\begin{align}
(x + i)(x-i) = x^2 - i^{2} = x^{2} - 1.
\end{align}
Use the Descartes Rule of signs to derive a contradiction. Hint: $i$ is not real. Can you finish the line of reasoning and derive an absurdity?
