Show that if $n$ and $m$ are both positive, then $nm$ is positive. This problem is on Terrence Tao's Analysis I, page 36.

Let $n,m$ be natural numbers. Then $n \times m=0$ if and only if atleast one of $n,m$ is equal to zero. In particular, if $n$ and $m$ are both positive, then $nm$ is also positive.

I know that multiplication is defined as recursive addition. $1 \times m := m$ and $(n+1) \times m= (n \times m) + m$. But, that's all that comes to my mind. I am not sure, how to find and organize my arguments to prove the above result. Should I use induction? Any hints, tips that lead me to prove this result would be helpful.
As a second question, what's a reasonable amount of time, I should spend on thinking about a theorem and sketching a proof, before I move on, perhaps try another question and then come back later? And should I search for a solution on the Internet, if I still don't get it.

Proof(Rough sketch).
Let $n,m$ be positive natural numbers. We induct on $n$ keeping $m$ fixed.
(I) Claim. $1 \times m$ is positive. 
$1 \times m := m$ by the definition of multiplication and $m$ is positive.
(II) We inductively assume that $n \times m$ is positive.
(III) We would like to show that $(n+1)\times m$ is positive.
By the definition of multiplication:
$\begin{aligned}
(n+1) \times m &= (n \times m) + m
\end{aligned}$
$n \times m$ is positive from the inductive assumption. $m$ is a positive natural number. The sum of two positive natural numbers is a positive number. This closes the induction.
