Square root of 1 is (not) -1

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$i^2$ why is it $-1$ when you can show it is $1$?

I was thinking on the following line of thoughts: $1 = \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$

Of course this is not true, but I was wondering which step in this 'line of thoughts' is forbidden to make?

Thanks for the explanation.

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$\sqrt{-1 \cdot -1}$ is not equal to $\sqrt{-1} \cdot \sqrt{-1}$. The formula $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is only valid when both $a,b$ are nonnegative real numbers.