Valid example of proof by contradiction?

I'm studying "How To Prove It" (Velleman) and I'm on exercise 3.7. The solution I gave is a "proof by contradiction" and I would like to verify that the method of proof and result are valid.

$\displaystyle \\\mbox{Suppose that }a\mbox{ is a real number. Prove that if }a^3 > a\mbox{ then }a^5 > a\mbox{.}\\\\\mbox{Suppose }a^5 \leq a\mbox{. Then }a^5 - a \leq 0\mbox{ and }a^5 - a = (a^3 - a)(a^2 + 1) \leq 0\mbox{.}\\\mbox{However, we know that }a^3 > a\mbox{ so the first factor }a^3 - a > 0\mbox{.}\\\mbox{The second factor is also positive since }x \ne 0 \rightarrow x^2 > 0\mbox{ for all }x \in \mathbb{R}\mbox{,}\\\mbox{which implies that }a^2 + 1 > 0\mbox{ as well. }\\\mbox{This leads to a contradiction however, therefore }a^3 > a \rightarrow a^5 > a\mbox{.}$

Is this a well formed proof? How explicit do I have to be? For example, is it OK for me to leave out the fact that for a, b in R, a > 0 and b > 0 implies that ab > 0? If so, then what else could I have left out?

Maybe these things will be more clear once working on more complicated proofs?