In a ring with identity, prove that (-1)(-1) = 1 I would like to attempt to solve this problem on my own if someone could give me a suggestion for how to start it.
 A: When you're just starting out with this stuff, I don't think you can do much better than trying out a bunch of different ways to write expressions involving $1$, $0$, $-1$, and the various arithmetic operations, and see if any of them look like they tell you anything interesting. (this includes both substituting values into identities to see if they're useful, and using identities to rewrite what you're trying to prove, to see if you can relate them to other things you can figure out more easily)
Hint: you really are going to have to make use of $+$ in this proof. (unless you've already proven a particular other identity that can be used to simplify this problem, anyways)
A: Hint  $\quad \overbrace{(-x)(-y)+ x(-y)}^{\large (-x+x)(-y) \,=\, 0}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\phantom{x(-y)}+xy}_{\large x(-y+y)\,=\,0}\ \Rightarrow\ (-x)(-y)\,=\,xy$
Said more conceptually, using the distributive law, etc $\,(-x)(-y)\,$ and $\,xy\,$ are inverses of $\,x(-y)\,$ hence they are equal by uniqueness of inverses, i.e. if $\,B\,$ has two additive inverses $\,A,A'\,$ so $ A+B = 0 = B+A'\,$ then $\, A' = (A+B)+A' = A+(B+A') = A.\,$ The hint is just a graphic way of evaluating $\,A'+B+A\,$ in two ways by reassociating the additions. 
