I have the statement $P(n)$: $2^n<(n+1)!$, for $n \geq 2$;
$P(2)$: $2^2 < 3!$ which is true
I.H P(k): $2^k<(k+1)!$
show that $P(k+1)$: $2^{k+1} <(k+2)!$
Here is my approach:
$2^k<(k+1)!$ multiply both sides by $2$
$2*2^k<2*(k+1)! \Rightarrow 2^{k+1} < 2*(k+1)!$
since: $(k+2)! = (k+2)*(k+1)! \Rightarrow k*(k+1)! + 2(k+1)!$
hence: $k*(k+1)! + 2(k+1)! > 2(k+1)! \Rightarrow (k+2)! > 2(k+1)!$
we can show that: $2^{k+1} < 2(k+1)! < (k+2)! \Rightarrow 2^{k+1} < (k+2)!$ for $n \geq 2$
I need a feedback about my approach and any correction in case of mistakes.
