How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$ 
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the right approach? Because when i try and work out the algebra i keep on hitting a wall. 
\begin{align}
\left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)&=1-\frac 1{(1-n)^2}-\frac 1{n^2}-\frac 1{n^2(n+1)} \\
&=\frac{n^2}{(n+1)^2}-\frac 1{(n+1)^2} \\
&=\frac{n^2-1}{(n+1)^2}-\frac 1{n^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)}{n^2(n+1)^2}-\frac{(n+1)^2}{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2}{n^2(n+1)^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2+1}{n^2(n+1)^2}
\end{align}

$\begin{aligned}&  \prod_{2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln{\prod_{2\le k\le n}\bigg(1-\frac{1}{k^{2}}\bigg)}\bigg] = \exp\bigg[\sum_{2 \le k \le n} \ln{\bigg(\frac{k^2-1}{k^2} \bigg)}\bigg] = \exp(S_{n}). \\& \begin{aligned} & \begin{aligned} S_{n} & = \sum_{2 \le k \le n}\ln(k-1)+ \sum_{2 \le k \le n}\ln(k+1)-2 \sum_{2 \le k \le n}\ln(k) =\sum_{2 \le k+1 \le n}\ln(k)+ \sum_{2 \le k-1 \le n}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) \\& = \sum_{1 \le k \le n-1}\ln(k)+ \sum_{3 \le k \le n+1}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) = \ln(1)-\ln(n)-\ln(2)+\ln(n+1)\\& +\sum_{2 \le k \le n}\ln(k)+ \sum_{2 \le k \le n}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) = \ln\bigg(\frac{1}{2}\bigg)+\ln\bigg(1+\frac{1}{n}\bigg) .\end{aligned} \\& \begin{aligned} \therefore \prod_{ 2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln\bigg(\frac{1}{2}\bigg)+\ln\bigg(1+\frac{1}{n}\bigg)\bigg]  = \frac{n+1}{2n}.  \end{aligned} \end{aligned} \end{aligned} $
