Comparing $2^n$ to $n!$

$$2^n \;=\;\; \underbrace{\;2\cdot 2 \cdot 2\cdot 2\cdot\, \cdots \,\cdot 2 \; \cdot \;2\;}_{\text{n $ $factors of $2$}}$$

$$n! = \underbrace{1\cdot 2\cdot 3 \cdot 4\cdot\, \cdots\,\cdot (n - 1)\cdot n}_{\text{n factors}}$$

When we compare the products $\,$ factor-by-factor,$\,$ we can easily see that when $n \geq 4$, $\;n!\;>\;2^n$.

This can, of course, be formalized using induction on $n$.

Consider for $n \ge 4$: $$ 2^n = 2 \cdot 2 \cdot 2^{n - 2} < 2 \cdot 3 \cdot 4 \cdot \ldots \cdot n = n! $$

By induction, $(n+1)!=(n+1) \cdot n! \geq (n+1) \cdot 2^n \geq 2 \cdot 2^n=2^{n+1}$ if $n! \geq 2^n$. Therefore, because $4$ is the smallest integer such that $2^n \leq n!$ is true, you deduce that $n! \geq 2^n$ for all $n \geq 4$.