Prove the following by using the principle of mathematical induction for all $n \in N$ where $n \geq 2$:
$\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right) \ldots \left(1-\frac{1}{n^{2}}\right)=\frac{n+1}{2n}$

Let $P(n)$ be the statement: $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \ldots \left(1-\frac{1}{n^{2}}\right)=\frac{n+1}{2n}$.
Step $1$: For $n=2$,the $LHS$ is $\left(1-\frac{1}{2^{2}}\right) = 1-\frac{1}{4} = \frac{3}{4}$.
The $RHS$ is $\frac{2+1}{2(2)} = \frac{3}{4}$.
Since $LHS$ = $RHS$,$P(2)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 2$,i.e.,$\left(1-\frac{1}{2^{2}}\right) \ldots \left(1-\frac{1}{k^{2}}\right) = \frac{k+1}{2k}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$\left(1-\frac{1}{2^{2}}\right) \ldots \left(1-\frac{1}{k^{2}}\right)\left(1-\frac{1}{(k+1)^{2}}\right) = \frac{(k+1)+1}{2(k+1)} = \frac{k+2}{2(k+1)}$.
Using the assumption from Step $2$:
$LHS$ = $\left(\frac{k+1}{2k}\right) \left(1-\frac{1}{(k+1)^{2}}\right) = \left(\frac{k+1}{2k}\right) \left(\frac{(k+1)^{2}-1}{(k+1)^{2}}\right)$.
$= \left(\frac{k+1}{2k}\right) \left(\frac{k^{2}+2k+1-1}{(k+1)^{2}}\right) = \left(\frac{k+1}{2k}\right) \left(\frac{k(k+2)}{(k+1)^{2}}\right)$.
$= \frac{k+2}{2(k+1)}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \geq 2$.