# Proof of square of -1

With a field axiom concerning a structure of real numbers $$\mathbb R$$, like the Closure, Associative, Commutative, Distributive law and existence of the identity, inverse, etc. How to prove that $$(-1)^2 = 1$$? Should proofs of all formulas or statements in this axiom start with "there exists the number $$1$$, $$0$$" when we can't use any preceding theorem?

$$(-1)^2-1=^{\color{red}{\vdash}}(-1)(-1+1)=0$$. Then $$(-1)^2=1$$ by the uniqueness of the neutral element.
$$\cdot$$ in $${\color{red}{\vdash}}$$ use distributive.
The existence of $$1$$ and $$0$$ are axioms too.
• $(-1)^2 + 1 \neq 0$; you mean $(-1)^2+(-1)$. – Arturo Magidin Mar 19 at 6:35
• $\left( -1 \right)^2 = -1$ ? Shouldnt it be $= 1$ ? – thinkingeye Mar 19 at 7:18