$$Lemma\text{:}\qquad \forall n\in \mathbb{R}, \ n^2 = (-n)^2 \ \because 1 = (-1)^2$$ $$\text{e.g.}\qquad \forall (x, y)\in \mathbb{R}, \ (x - y)^2 = \big(-(y - x)\big)^2 : (x - y)\in \mathbb{R}\ \land \ (x - y) = -(y - x)$$. $$\begin{align} \text{Pursuant to the Lemma, }\ \sqrt{-1}^2 &= (-\sqrt{-1})^2 \\ &= \big((-1)\sqrt{-1}\big)^2 \\ &= (\sqrt{-1}^2\sqrt{-1})^2 \\ &= (\sqrt{-1}^3)^2 \\ &= \sqrt{-1}^{3\times 2} \\ &= \sqrt{-1}^6 \\ \end{align}$$. $$\therefore \sqrt{-1}^2 = \sqrt{-1}^6$$ I know that with the rules of $\sqrt{-1}$, this is allowed. But doesn't this break the rules of the Lemma since $\mathbb{R} \subseteq \mathbb{C}$ and $\sqrt{-1}\in \mathbb{C}?$ The only difference this would be proposing in the Lemma would be that it is false for all $n\in \mathbb{R}$.
$(Lemma = \text{a subsidiary proposition; a helping theorem})$
My Attempt: $$\begin{align} \sqrt{-1}^2 &= (-\sqrt{-1})^2 \\ &= \big((-1)\sqrt{-1}\big)^2 \\ &= (-1)^2\sqrt{-1}^2 \\ &= 1\times \sqrt{-1}^2 \\ &= \sqrt{-1}^2 \\ &= -1 \\ \end{align}$$ This way, it seems like the Lemma must be true, but in the same way, we can write: $$\begin{align} \sqrt{-1} &= (\sqrt{-1}^2)^2\sqrt{-1}^2 \\ &= \sqrt{-1}^4\sqrt{-1}^2 \\ &= \sqrt{-1}^{4 + 2} \\ &= \sqrt{-1}^6 \\ \end{align}$$
Could you please explain to me what I am doing incorrectly (if I am, or if not, what I am interpreting incorrectly)? I only know the "algebra" part of imaginary numbers and not the "geometric" part. If you are using different techniques to explain the reasoning behind this, please be clear and show me step by step.
Thank you in advance.
$$(-n)^2 = (-n).(-n) = -1.n.-1.n = (-1).(-1).n.n = ((-1).(-1)).(n.n) = 1.n^2 = n^2$$