The real question is why negative times negative is positive - I had this doubt in school. I am a bit ashamed to admit that I still don't know any "from first principle explanation" for this fact.

Don't sweat it, here's a rough proof. First a proof that a negative times a positive is negative:

    a*0 = 0 [Assumed but you can prove this too.]
    a*(b-b) = 0 [Definition of 0.]
    a*b + a*(-b) = 0 [Distributive law.]

For any X, only its inverse -X has the property that X + -X = 0. Therefore a×(-b) is the inverse of a×b [which is -(a×b)].

Then apply the same idea but with a negative a:

    (-a)*0 = 0 [Assumed but you can prove this too.]
    (-a)*(b-b) = 0 [Definition of 0]
    (-a)*b + (-a)*(-b) = 0 [Distributive law.]
    -(a*b) + (-a)*(-b) = 0 [From above.]
    (-a)*(-b) + -(a*b) = 0 [Commutative addition to rearrange to X + -X = 0 form.]

And as before, only the inverse has this property so -(a×b) must be the inverse of (-a)×(-b) -- which is a×b! So a negative number times a negative number is positive. QED.

In case you're interested, the proof that the additive element is a multiplicative absorbing element (a×0 = 0, for all a) is also fairly trivial[1].

The above does somewhat implicitly assume that 0 and -0 are equal (or at least that (-a)×0 = (a)×0 = 0, which implies that 0 and -0 are equal).

[1]: https://en.wikipedia.org/wiki/Additive_identity#The_additive...