For $n=4$, we have that $4!=24>16=2^4$, so the condition holds. Now assume it's valid for $k\geq4$: $k!>2^k$. Then $$(k+1)!=(k+1)k!>(k+1)2^k>2\cdot2^k=2^{k+1},$$ and we hake that the condition holds for $k+1$.
For $n=4$, we have that $4!=24>16=2^4$, so the condition holds. Now assume it's valid for $k\geq4$: $k!>2^k$. Then $$(k+1)!=(k+1)k!>(k+1)2^k>2\cdot2^k=2^{k+1},$$ and we hake that the condition holds for $k+1$.