TITLE: Is this a proper proof of (-1)(-1) = 1?
QUESTION [8 upvotes]: I am a novice in proof writing and have just started a book on analysis. I have no other pure math experience or knowledge of abstract algebra. 
I am trying to prove that $(-1)*(-1) = 1$. I will first show my attempt and follow it with the standard technique that I have found online. I would like to understand why the second method is preferred (or perhaps the only valid one).
Method 1 (my attempt): We are given the nine field axioms for $\Bbb R$. Using these, it has been shown (in the text) that $(-1)x = -x$ for all $x \in \Bbb R$. Therefore, $(-1)(-1) = -(-1) = 1$ because we have the identity element of multiplication $1 \in \Bbb R$ and there exists $ -1 \in \Bbb R$ such that $-1$ is the additive inverse of $1$. 
Method 2 (found by Google search): We know that $(0)(0) = 0$ because it was shown (in the text) that $0x = 0$ for all $x \in \Bbb R$. Then $$(0)(0) = ((-1) + 1)*((-1) + 1) = (-1)(-1) + (-1)(1) + (1)(-1) + (1)(1) \\= (-1)(-1) - 1 - 1 + 1 = (-1)(-1) - 1 = 0$$ Therefore $(-1)(-1) = 1$ because we see that $(-1)(-1)$ is the additive inverse of $-1$. 
Is the second method simply a more precise version of the first method, while the first is overly verbal? Or is the first somehow using circular reasoning?

REPLY [8 votes]: I'd do it directly as follows:
$$(-1)(-1)+(-1)\stackrel{Dist.}=(-1)\left[(-1)+1\right]\stackrel{\text{additive inv.}}=(-1)\cdot 0\stackrel{text}=0$$
Thus, $\;-1\;$ is the additive inverse of $\;(-1)(-1)\;$ , but also $\;-(-1)=1\;$ is the additive inverse of $\;(-1)\;$, so by uniqueness of the inverses we're done.