Assume that by $i$ we mean $\sqrt{-1}$. We know that $$ 1-i=\sqrt{2}\, e^{{\Large\frac{7\pi}{4}i}} \tag{1} $$
Now I want to obtain the value of $1-i$ by elementary mathematics, but I do not know where I made a mistake such that the final result does not match with ($1$). $$ 1-i=\sqrt{(1-i)^2}=\sqrt{1-1-2i}=\sqrt{-2i}=\sqrt{2}\sqrt{-1}\sqrt{i} =\sqrt{2}ii^{1/2}=\sqrt{2}i^{3/2} $$ Moreover, we know that $i=e^{\Large{\frac{\pi}{2}i}}$. Therefore, we get $$ 1-i=\sqrt{2}\, e^{\Large{\frac{3\pi}{4}i}} \tag{2} $$
$$\sqrt{(1-i)^2}=\pm(1-i)=\begin{cases}\sqrt2e^{3\pi i/4}&\text{if }-\\\sqrt2 e^{7\pi i/4}&\text{if }+\end{cases}$$ Your first step does not hold, $(1-i)$ is not the only square root of $(1-i)^2$. By similar reasoning, we could get, say $$-i=\sqrt{(-i)^2}=\sqrt{-1}=i$$