= Solution
\b[False.] It suffices to find a subsequence along which the ratio is unbounded. Set $n=4m$. The last $3m$ factors of $(4m)!$ are at least $m$, while the first $m$ factors have product at least one. Hence
$$
\frac{(4m)!}{(4m)^{2m}}
\ge\frac{m^{3m}}{16^m m^{2m}}
=\left(\frac m{16}\right)^m\longrightarrow\infty.
$$
This contradicts the eventual uniform bound required by <Big O notation>. The proof is an elementary <factorial> comparison, <independent> of <Stirling's approximation>.
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