Induction proof verification ##2^{n+2} < (n+1)## for all n ##\geq 6##

[ciencero]

$2^{n+2} < (n+1)!$ for all n $\geq 6$

Step 1: For n = 6,

$256 < 5040$.

We assume

$2^{k+2} < (k+1)!$

Induction step:

$2 * 2^{k+2} < 2*(k+1)!$

By noting $2*(k+1)! < (k+2)!$

Then $2^{k+3} < (k+2)!$

 

[mfb]

Looks fine.
n=6 is not the first place where the inequality is true, by the way.

 

[Mark44]

$2^{n+2} < (n+1)!$ for all n $\geq 6$

Step 1: For n = 6,

$256 < 5040$.

We assume

$2^{k+2} < (k+1)!$

Induction step:

$2 * 2^{k+2} < 2*(k+1)!$

By noting $2*(k+1)! < (k+2)!$

Then $2^{k+3} < (k+2)!$

@ciencero, at this site, use double $ characters at each end for standalone LaTeX, or double # characters at each end for inline LaTeX.

 

[WWGD]

It seems clear the RH side will eventually dominate. LH is being multiplied by 2 from nth to (n+1)st term while RH side is being multiplied by increasingly larger factors.