# $(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$. (Mathematical Induction) [closed]

By using Mathematical Induction prove that $$(n+1)!>2^{(n+1)}$$ for $$n$$, where $$n \geq 4$$.

First, check the base case: $$((4)+1)!=120>32=2^{(4)+1}$$ Next we want to show that $$(k+1)!>2^{k+1}\implies (k+2)!>2^{k+2}$$. Since $$(k+2)>2$$ for all $$k \geq 4$$ and by hypothesis $$(k+1)!>2^{k+1}$$, we get $$(k+2)!=(k+2)(k+1)!>2\cdot 2^{k+1}=2^{k+2}$$
If true for some $$k\geq{4}$$, that is $$(k+1)!>2^{k+1}$$, then $$(k+2)!=(k+2).(k+1)!>(k+2).2^{k+1}>2.2^{k+1}=2^{k+2}$$
$$n!=\overbrace{1\times 2\times 3\times 4}^{=24\, >\, 2^4}\times \overbrace{5}^{>2}\times \overbrace{6}^{>2}\cdots\times \overbrace{n}^{>2}>2^4\times 2^{n-4}$$