Okay, I've dug out my old Real Analysis textbook and I have the proof here. I'll try to (loosely) translate from "math speak" into English. First of all, some background: the axioms of the real number system consist solely of the algebraic properties and the order properties. The algebraic properties are common to other fields and include commutativity and associativity for both addition and multiplication, the existence of both a multiplicative identity (1) and an additive identity (0), inverse elements for addition and multiplication (i.e. negative numbers and reciprocals), and the distributive property. The order properties state that the sum of 2 positive numbers is a positive number, the product of 2 positive numbers is a positive number, and finally any real number is either positive, negative, or 0 (the additive identity element).
All other facts about real numbers can be derived from these. To prove 1 > 0, first we have to show that the square of any nonzero number is postive (you'll see why this is important in a bit). This follows from the order properties: if a is positive then (a)(a) is also positive, and if a is negative, then since (a)(a) = (-a)(-a), and -a is postive, (a)(a) is still positive.
Okay, now the key part: since 1 is the identity element, we know 1 times 1 = 1. So 1 must be greater than 0.
Having written the above, I now realize that this proof (that 1 > 0) actually presupposes that 1 does not equal 0, so actually this doesn't address Ramned's original question at all. Oops. Still, some might find the above interesting so I'll still post it. In fact, looking back at the axioms, it states specifically that the multiplicative identity (1) is distinct from the additive identity (0).
Incidentally, once you've proven 1 > 0, you can then prove that any natural number is greater than 0 by induction (we know 1 > 0, and by the order properties if k is positive k + 1 must also be positive).