Recent content by 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 $2^{k+3} < (k+2)!$