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

$2^{n+2} < (n+1)!$ 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 $2^{k+3} < (k+2)!$

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

Mark44
Mentor
$2^{n+2} < (n+1)!$ 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.

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