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}

Check that the equation holds for $n=2$. Assuming that it holds for some $n$:
$$\begin{align}\prod_{i=2}^n\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\\
\end{align}$$
And so
$$\begin{align}
\prod_{i=2}^{n+1}\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\left( 1-\frac1{(n+1)^2}\right)\\
&=\frac{n+1}{2n}-\frac{n+1}{2n(n+1)^2}\\
&=\frac{(n+1)^2}{2n(n+1)}-\frac{1}{2n(n+1)}\\
&=\frac{(n+1)^2-1}{2n(n+1)}\\
&=\frac{n^2+2n}{2n(n+1)}\\
&=\frac{n+2}{2(n+1)}\\
\end{align}$$
If the equation holds for some $n$, it also holds for $n+1$. Since it holds for $2$, it also holds for $3$, and since it holds for $3$, it also holds for $4$, and so on. The statement is true for all natural numbers greater than or equal to $2$.
