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)