Okay, so by the definition of iota we know it is equal to $\sqrt{-1}$.
(1st STEP). $i^2 = (\sqrt i)^2$
(2nd STEP). $i^2 = \sqrt{-1} \times \sqrt{-1}$
( I did not $\sqrt{I}$ directly as I wanted to understand the squaring process of iota in greater depth)
Now, I also cannot use the identity $\sqrt a \times \sqrt b = \sqrt{ab}$ because both $a$ and $b$ are negative in this problem. So, I am confused as to how I should proceed further using normal arithmetic.
By very definition, $i^2=-1$. Writing $i=\sqrt{-1}$ may be a way to think about the nature of $i$, but it must be banned whenever one does arithmetic operations on complex numbers, because of this confusion about how to handle the $\sqrt{\;}$ sign. One should always use the notation $i$ (or $j$ for engineers) along with the simple rule that $i^2=-1$.
By the way, this is the letter i, $i$. Not iota !