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