Why is $i^3$ (the complex number "$i$") equal to $-i$ instead of $i$? $$i^3=iii=\sqrt{-1}\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i
$$

Please take a look at the equation above. What am I doing wrong to understand $i^3 = i$, not $-i$?
Since $i^2=-1$ by definition, $i^3=i^2\cdot i=-i$.

$\sqrt{a}\sqrt{b}=\sqrt{ab}$ is only guaranteed for positive real $a$ and $b$.
Someone posted earlier that it is i x i x i which is -1 x i, ie -i (but the post seems to have been deleted!

But that's all there is to it. It doesn't matter what i represents, the algebra will be consistent. But to prove it, substituting the value of i, ie (√-1), you get

√-1 x √-1 x √-1 
= -1 x √-1
= -(√-1)
= -i

However it may be worth mentioning that i is not a complex number as has been suggested - it is an irrational number. A complex number is of the form (x +iy)