Proof by Induction: $n! > 2^{n+1}$ for all integers $n \geq 5.$

Question

I have to answer this question for my math class and am having a little trouble with it.

Use mathematical induction to prove that $n! > 2^{n+1}$ for all integers $n \geq 5.$

For the basis step: $(n = 5)$

$5! = 120$

$2^{5+1} = 2^6 = 64$

So $120 > 64$, which is true.

For the induction step, this is as far as I've gotten:

Prove that $(n! > 2^{n+1}) \rightarrow \left((n+1)! > 2^{(n+1)+1}\right)$

Assume $n! > 2^{(n+1)}$ Then $(n+1)! = (n+1) \cdot n!$

After this, I'm stuck. Any assistance would be appreciated.

Thanks!

Answer 1

$(n+1)!=(n+1)\times n!>(n+1)2^{n+1}>2\times2^{n+1}=2^{n+2}$

Make sure you understand all the steps and ask if you got trouble.

Answer 2

Induction step:

\begin{equation} n! > 2^{n+1} \implies (n+1)! = n!(n+1) > 2^{n+1}(n+1)>2^{n+1}\;·\;2 = 2^{n+1+1} = 2^{n+2}. \end{equation} The last inequality is because $n\geq 5$ implies $n+1 \geq 6 > 2$.

Answer 3

If $n > 1$ and $n! > 2^{n+1} $ then

$(n+1)! = (n+1)n! > (1+1)n!=2*n! > 2*2^{n+1} = 2^{n+2} $

...

We just need the base step to establish for $n=5$ that i) n=5 > 1 (duh). And ii) $n! = 5!=120$ is larger than $2^{n+1}=2^6=64! $