$n!$ vs $2^n$

Base case. $n=4$: $4!=24>16=2^4$. ✓

Inductive step. Assume $n!>2^n$ with $n\ge4$. Then
$$(n+1)!=(n+1)\cdot n!>(n+1)2^n>2\cdot2^n=2^{n+1},$$
since $n+1>2$.

Conclusion. $n!>2^n$ for all $n\ge4$.