Mistake !!

I'm going to prove that $1=-1$ using some flawed logic in the following steps.

Step 1: It is true that $\begin{aligned} 2 & = \dfrac{1}{1-\frac{1}{2}} ={\color{#3D99F6} \dfrac{1}{1\times \frac{1}{2}}}= 2\,.\end{aligned}$

Step 2: Let's take one of two and make substitution of 2 by next

$\begin{aligned}2 & = \dfrac{1}{1-\frac{1}{2}}= \dfrac{1}{1- \dfrac{1}{{\color{#3D99F6}\frac{1}{1\times \frac{1}{2}}}}}\,.\end{aligned}$

Step 3 : Solving the expression as $\begin{aligned}2 = \dfrac{1}{\dfrac{\left({\color{#3D99F6}\dfrac{1}{1\times\frac{1}{2}}}-1\right)}{\dfrac{1}{1\times \frac{1}{2}}}}= \dfrac{\dfrac{1}{1\times \frac{ 1}{2}}}{ \dfrac{1-\frac{1}{2}}{\frac{1}{2}}}\,.\end{aligned}$

Step:4 With further simplication $\begin{aligned}2 &= \dfrac{2}{\dfrac{2-1}{\frac{1}{2}\times\frac{1}{2}}} = 2^{-1} \,.\end{aligned}$

Step 5 : It is proved that $2 = 2^{-1}$ and hence $1 =-1$ .

In which of the steps above , I make first mistake .

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This problem is original.