Comparing $2^n$ to $n!$
Solution 1:
$$2^n \;=\;\; \underbrace{\;2\cdot 2 \cdot 2\cdot 2\cdot\, \cdots \,\cdot 2 \; \cdot \;2\;}_{\text{n $ $factors of $2$}}$$
$$n! = \underbrace{1\cdot 2\cdot 3 \cdot 4\cdot\, \cdots\,\cdot (n - 1)\cdot n}_{\text{n factors}}$$
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 $n$.
Solution 2:
Consider for $n \ge 4$: $$ 2^n = 2 \cdot 2 \cdot 2^{n - 2} < 2 \cdot 3 \cdot 4 \cdot \ldots \cdot n = n! $$