Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. For $n=0$, we have $LHS=2^0=1$, and $RHS=(0+2)!=2$, thus $LHS\lt RHS$ and $P(0)$ is true.
3. Assume $P(k), k\gt0$ is true, that is $2^k\lt (k+2)!$.
4. For $n=k+1$, we have $LHS=2^{k+1}\lt 2\cdot (k+2)!\lt (k+3)(k+2)!=(k+3)!=RHS$
5. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.