Which step in this process allows me to erroneously conclude that $i = 1$

Exposing the scam becomes easier if we run an analogous one in the reals:

$-1 = (-1)^1$
$\phantom{-1} = (-1)^{2/2}$
$\phantom{-1} = ((-1)^2)^{1/2}$
$\phantom{-1} = (1)^{1/2}$
$\phantom{-1} = 1$

One can now see the deception, going from the second line to the third: if we've decided to interpret $x^{1/2}$ always as positive square root (or indeed if we've made any choice of square roots in order to regard $x \mapsto x^{1/2}$ as a function), then $x^{2/2} = (x^2)^{1/2}$ will not necessarily hold.

It is true that $i$ and $1$ are both fourth roots of $1$, just not the same fourth root. You have chosen a branch of the fourth root function, and you have assumed algebraic properties of it that don't exist. You need to define $x^{1/4}$, $\sqrt x$, etc.; there is no way to do so that is consistent with your steps. In general, don't assume $x^{ab}=(x^a)^b$ without reason.

Already with $i=\sqrt{-1}$ I am suspicious. There is more than one squareroot of $-1$, and there can be trouble with that notation if you aren't careful. Rewriting that as $(-1)^{1/2}$ doesn't really change anything, but changing the exponent to $2/4$ indicates more trouble as you get ready to take fourth roots. There are 4 complex fourth roots of a nonzero complex number, and you are later going to make a choice about which to take. The step where $i$ becomes $((-1)^2)^{1/4}$ is definitely trouble; $(x^2)^{1/4}=x^{1/2}$ is not an identity for complex numbers. It will work sometimes, depending on what branches of the square root and fourth root functions you take and what $x$ is, and other times it will not work. You can think of your work as a proof that it does not hold with your choice of $(-1)^{1/2}=i$ and $1^{1/4}=1$.

The general rule $\left(x^a\right)^b = x^{a\cdot b}$ stands only for $\left(a,b\right) \in \mathbb{Z}^2$.

Also the notation $\sqrt{ \times }$ is abusive and should not be used unless speaking of positive real numbers, because each complex number has two square roots (0 has only one) but for positive reals we can arbitrarily decide that this notation maps to the real positive one.

Replace every mention (outside the exponent) of $i$, $-1$ and $1$ respectively by $X$, $X^2$ and $X^4$. This does not involve any ambiguity about signs or square roots or branch cuts, and uses only the always valid identities $i^2 = -1$, $(-1)^2 = 1$, and $(a^2)^2 = a^4$. The argument now reads:

$ X = \sqrt{X^2} $
$ \sqrt{X^2} = (X^2)^{1/2} $
$ (X^2)^{1/2} = (X^2)^{2/4} $
$ (X^2)^{2/4} = ((X^2)^{2})^{1/4} $
$ ((X^2)^{2})^{1/4} = (X^4)^{1/4} $
$ (X^4)^{1/4} = 1 $

Some vertical equality signs are implied but not written.

All the equalities are either correct, or deduced from an axiom that $(T^{n})^{1/n} = T = (T^{1/n})^{n}$.
In each use of the axiom, $T$ is $X$, $X^2$ or $X^4$ and $n$ is $1,2$ or $4$.

The second equation in the axiom is correct. The first is not. There are difficulties in consistently choosing the sign of the square root, when working with complex numbers, and the first equation (in effect) pretends that a consistent choice of $n$-th root can always be made.