Product of two natural numbers is also natural.

Yes, this prove is correct. Make sure to mention in the beginning that you pick $m$ as an arbitrary natural number. Since then you get the result that the multiplication of ANY two natural numbers is a natural number. Also don't forget to address the zero element (if you consider that to be a natural number).

Well, one can define addition and multiplication as follows:

$m+0=m, ~m\cdot 1 = m$

$m+n' = (m+n)', ~m\cdot n' = m\cdot n+m$

where $n'$ is the successor of $n$ defined by the injective mapping $n\mapsto n'$ with $0$ not in the image set.

Using induction on $n$, it's clear that both operations are well-defined.