Solution
= Solution
Prove the explicit bound $n!\leq2^{n^2}$ by <mathematical induction>. At $n=1$, $1\leq2$. For the step, note that $n+1\leq2^n$ for $n\geq1$: this auxiliary inequality has base $2=2$ and follows at the next index from $n+2\leq2(n+1)$. Hence, using the <induction hypothesis>,
$$
(n+1)!\leq(n+1)2^{n^2}\leq2^{n^2+n}\leq2^{(n+1)^2}.
$$
Thus \b[$n!=O((2^n)^n)$], with constant $1$ and threshold $1$ in <big O notation>.