Show that $2^n=O(n!)$
Proof:
By definition of Big-O, $\exists$ constants $c$ and $n_0$ such that $2^n \le cn!$ $\forall $ $n \ge n_0$. For a large $n$, since $2^n = \underbrace{2 \cdot 2 \cdot 2\cdot\cdot\cdot2}_{n\text{ times}}$ and $n! = 1 \cdot 2 \cdot 3 \cdot\cdot\cdot n$ Clearly, as $n \rightarrow +\infty$ , $2^n \le cn!$.
Alternatively, WLOG, we use induction and show that $2^{n+1} \le c(n+1)!$.
$2^n \le cn!$ $2^{n+1} \le 2(cn!) \le (cn+1)! \le c(n+1)! $ since $cn+1 \ge 2.$
Question: Is my proof correct? and is there a better way of proving such. thanks