Assume $n$ is much greater than $m$. Would $O(nm)$ simplify to $O(n)$? Is there an explanation?
Simplified big O notation of $O(nm)$ for $n\gg m$
asymptotics
It depends what $n$ is much greater than $m$ means but, in general, no. For example, one common definition of much greater is that $m=o(n)$. But, in this case, we can still have, e.g., $m=n^\epsilon$ for some $\epsilon<1$, or $m=\log^c n$ for any $c\geq 0$ and, in both of these cases, $nm\neq O(n)$.To get $nm=O(n)$, you need $m=O(1)$.