Solution
= Solution
\b[False.] Set $r_n=n!/2^n$. The ratio of successive terms is
$$
\frac{r_{n+1}}{r_n}=\frac{n+1}{2}\geq2\qquad(n\geq3).
$$
Consequently $r_n\geq r_3\,2^{n-3}$ for $n\geq3$, which is unbounded. No constant can bound $n!$ by that constant times $2^n$ for all sufficiently large $n$. The <factorial> therefore fails the proposed <Big O notation> bound, without needing <Stirling's formula>.