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}

Some of the confusion here is because there are two types of roots: roots and principal roots. The principal roots always return only one value. However symbolically they are almost never different. You can very, very rarely encounter $_+\sqrt{\;}\;$for the principal root. So confusion is often created because people confuse these two types of roots, as they are represented by exactly the same symbols. 
$$_+\sqrt1=1$$
$$\sqrt1=-1,+1$$
$$\sqrt{-1}=-i,+j$$
If we are doing something like this
$$\sqrt{-1}=\frac{1}{i}$$ it means we have to take the correct roots; otherwise, we can have even this:
$$\sqrt{1}=-1\;\ and \;\sqrt{1}=1$$
Hence, $1=-1\,.\;$
That's not the way to go. We have multiple roots when dealing with complex numbers, even square roots from real numbers have two answers. To avoid this, the principal root is used but it doesn't differ symbolically from just square root. People extremely rarely write $_+\sqrt{\;}\;$. Therefore there's a lot of confusion. The rules of extracting roots from complex numbers don't strictly follow the rules used with principal roots, so you may easily arrive at a wrong answer. So when we see, for example, the formula:
$$\sqrt[n]{z}=\sqrt[n]{p}\, \bigg(\cos \frac{φ+2πk}{n}+i\sin\frac{φ+2πk}{n}\bigg)$$ 
We have to understand that what is meant is this
$$\sqrt[n]{z}=\,_+\sqrt[n]{p}\, \bigg(\cos \frac{φ+2πk}{n}+i\sin{\frac{φ+2πk}{n}}\bigg)$$ 
Another example. We can't reduce a multiple root $\;\sqrt[nk]{z^k}\;$ to $\;\sqrt[n]{z}\;$ because the first one has $nk$ different root values and the second--only $n\,$.
Formulas such as $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ will work but not always. Generally, we can't use rules for principal roots for real numbers when we are dealing with complex numbers.
To recap:
First, $\sqrt{-1}=-i,+i\;\;$ (   NOT JUST $\;i\,$)
Second, when dealing with complex numbers, $\sqrt1=-1,+1\;\;$ ( NOT JUST $\;1\;$)
And third, $\sqrt{-1}\cdot\sqrt{-1}=1\;$ if we take $\;i,-i\;$ or $\;-i,i\;$ as roots, and  $\sqrt{-1}\cdot\sqrt{-1}=-1\;$ if we take either $\;i,i$ or $\;-i,-i\;$ as roots. And even if we deal with real numbers and extract roots (not principal roots) we can still arrive at different values.
In very old books $\sqrt{-1}$ is sometimes used instead of $i$. It can only add to confusion. 
Most crucial here: Roots must be distinguished from principal roots. When dealing with complex numbers we don't extract some principal root of them, but we have a set of different root values. So, almost every single line is wrong in the OP's 'proof'. Kevin Holt provided a very nice step by step illustration.
