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

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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 $$(k+2)!=(k+2)(k+1)!>2\cdot 2^{k+1}=2^{k+2}$$

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If true for some $k\geq{4}$, that is $(k+1)!>2^{k+1}$, then $(k+2)!=(k+2).(k+1)!>(k+2).2^{k+1}>2.2^{k+1}=2^{k+2}$

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$n!=\overbrace{1\times 2\times 3\times 4}^{=24\, >\, 2^4}\times \overbrace{5}^{>2}\times \overbrace{6}^{>2}\cdots\times \overbrace{n}^{>2}>2^4\times 2^{n-4}$

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