Let $S(n)$ be the statement : $2n< (n+1)!$ $n\geq 2$
Basic Step $S(1)$ :
LHS$=2*2=4$
RHS$=(2+1)!=6$
$LHS< RHS$ (verified).
Inductive Step :
Let us assume $S(k)$ is true .
So, $2k< (k+1)!$
Now,$S(k+1)$:
LHS::-
$2(k+1)$
$=2k+2< (k+1)!+2<(k+1)!*(k+2)<(k+2)!$
So,$S(k+1)$ is true whenever $S(k)$ is true.
So, $2n<(n+1)!;$$n\geq 2$ (proved)