Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? I know there must be something unmathematical in the following but I don't know where it is:
\begin{align}
\sqrt{-1} &= i \\\\\
\frac1{\sqrt{-1}} &= \frac1i \\\\ 
\frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\ 
\sqrt{\frac1{-1}} &= \frac1i \\\\ 
\sqrt{\frac{-1}1} &= \frac1i \\\\ 
\sqrt{-1} &= \frac1i \\\\ 
i &= \frac1i \\\\ 
i^2 &= 1 \\\\ 
-1 &= 1 \quad !!?
\end{align}

Isaac's answer is correct, but it can be hard to see if you don't have a strong knowledge of your laws. These problems are generally easy to solve if you examine it line by line and simplify both sides.
$$\begin{align*}
\sqrt{-1} &= i &
\mathrm{LHS}&=i, \mathrm{RHS}=i
\\
1/\sqrt{-1} &= 1/i &
\mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i
\\
\sqrt{1}/\sqrt{-1} &= 1/i &
\mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i
\\
\textstyle\sqrt{1/-1} &= 1/i &
\mathrm{LHS}&=\sqrt{-1}=i, \mathrm{RHS}=-i
\end{align*}$$
We can then see that the error must be assuming $\textstyle\sqrt{1}/\sqrt{-1}=\sqrt{1/-1}$.
