algebra precalculus - 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 $i^2$.

Answer

From $i^2=-1$ you cannot conclude that $i=\sqrt{-1}$, just like from $(-2)^2 = 4$ you cannot conclude that $-2=\sqrt 4$. The symbol $\sqrt a$ is by definition the positive square root of $a$ and is only defined for $a\ge0$.

It is said that even Euler got confused with $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$. Or did he? See Euler's "mistake''? The radical product rule in historical perspective (Amer. Math. Monthly 114 (2007), no. 4, 273–285).