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