Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. For $n=2$, we have $LHS=2^2=4$, and $RHS=(2+1)!=6$, thus $LHS\lt RHS$ and $P(2)$ is true.
3. Assume $P(k), k\gt2$ is true, that is $2^k\lt (k+1)!$.
4. For $n=k+1$, we have $LHS=2^{k+1}=2\cdot 2^k\lt 2\cdot (k+1)!\lt (k+2)\cdot (k+1)!= (k+2)!=RHS$.
5. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.