$\begin{aligned} \sqrt{-1}=&i \quad&\ldots(1) \\ \dfrac{1}{\sqrt{-1}}=&\dfrac{1}{i}\quad& \ldots(2) \\ \dfrac{\sqrt {1}}{\sqrt{-1}}=&\dfrac{1}{i} \quad &\ldots(3) \\ \sqrt{\dfrac{1}{-1}}=&\dfrac{1}{i} \quad &\ldots(4) \\ \sqrt{\dfrac{-1}{1}}=&\dfrac{1}{i}\quad &\ldots(5) \\ \dfrac{\sqrt{-1}}{1}=&\dfrac{1}{i} \quad &\ldots(6) \\ i=&\dfrac{1}{i} \quad& \ldots(7) \\ i^2=&1 \quad & \ldots (8) \\ -1=&1\quad & \ldots(9) \end{aligned}$

Consider these steps above.

In which step is the (first) mistake committed?