I need to use mathematical induction to prove $\displaystyle 2^n<n!$ for all n>=4. The base case is easy, and in the inductive step I get to $\displaystyle 2(2^n)<(n+1)n!$ But am I going in the wrong direction, or am I just missing something?
I need to use mathematical induction to prove $\displaystyle 2^n<n!$ for all n>=4. The base case is easy, and in the inductive step I get to $\displaystyle 2(2^n)<(n+1)n!$ But am I going in the wrong direction, or am I just missing something?
Assume $\displaystyle 2^k < k! $ for $\displaystyle k \geq 4 $. Then $\displaystyle 2^{k+1} < 2k! < (k+1)! = k!(k+1) $. $\displaystyle k+1 > 2 $ for $\displaystyle k >1 $. Hence $\displaystyle 2^{k+1} < (k+1)! $ for $\displaystyle k \geq 4 $ and we have proved the inductive step.