Value of $i^2$ in complex numbers

Please solve this doubt : we know that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and $i^2 = -1$. But $i= \sqrt{-1}$ which implies that $i^2 = i \cdot i = \sqrt{-1}\sqrt{-1} = \sqrt{1} = 1$ that is $i^2 = 1$. So what is the correct value of $i^2$? Is it $1$ or is it $-1$? Please explain.