Not defining the imaginary number $i$ as the principal square root of $-1$. Background
I learned early on that it's important that we define the imaginary number $i$ such that $i^2 = -1$, rather than $i = \sqrt{-1}$.
Question
I can't fully remember the reasoning for this important note, so I was wondering if someone could elaborate?
Own efforts
Any explanation I find eventually boils down to the same argument.

If we define $i$ as the principal square root of $-1$, then we get
$$-1 = i^2 = \sqrt{-1}\sqrt{-1} \overbrace{=}^{\text{fallacy}} \sqrt{(-1)(-1)} = \sqrt1 = 1$$

But to  me, this seems like wrongful use of the $\sqrt{ab} = \sqrt a \sqrt b$ rule, since this rule comes with certain restrictions on $a, b$. So I don't see how this is a misuse of the definition of $i$.
Are there other reasons why we should be careful not to define $i$ as the principal square root of $-1$?
 A: There are obviously two square roots of $-1$ (I'll assume there is a square-root of $-1$, such that that constructed from field extension), let's write them as $\pm\sqrt{-1}_{me}$.  And each person, knowing only arithmetics on $\mathbb{R}$, is free to choose his/her $\sqrt{-1}_{person}\in\{\pm\sqrt{-1}_{me}\}$, so there is no guarantee that $\sqrt{-1}_{me}=\sqrt{-1}_{you}$.  This is rather undesirable, as we never actually managed to pin down exactly what we meant by $i=\sqrt{-1}$.
A: If you define $i$ as $\sqrt{-1}$ then there is an obvious question: how do you know that $-1$ has some square root? Besides, writing $i=\sqrt{-1}$ seems to imply that $i$ is the square root of $-1$. But, in $\mathbb C$, $-1$ has two square roots: $\pm i$. Assuming that $i$ is the square root of $-1$ leads to fallacies, such as the one the you mentioned.
A: Well, one can define the complex numbers as ${\Bbb C} = \{a+ib\mid {\Bbb R}\}$, where $a+ib$ are considered as formal sums. Then one only needs that $i^2=-1$ for defining the multiplication.
The addition is componentwise,
$$(a+ib) + (c+id) = (a+c) + i(b+d)$$
and the multiplication (by multiplying term by term) is
$$(a+ib)\cdot (c+id) = ac + iad + ibc + i^2bc = (ac-bd) + i(ad+bc),$$
where $i^2=-1$ is used but not the square root.
Moreover, algebraically, $i^2=-1$ implies that $i$ is a primitive fourth root of unity, and, graphically, the numbers $a+ib$ can be drawn as usual in the Gaussian plane. No need of the square root here.
