Simplifying the product $\prod\limits_{k=2}^n \left(1-\frac1{k^2}\right)$ [duplicate]

Can we simplify the given product to a general law? $$\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{n^2}\right)$$

We can write, from the difference of squares, \begin{gather*} \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dots\left(1-\frac{1}{n^2}\right)\\ =\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\dots\left(1-\frac{1}{n}\right)\cdot \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{n}\right)\\ =\frac{1}{2}\frac{2}{3}\dots\frac{n-1}{n}\cdot\frac{3}{2}\frac{4}{3}\dots\frac{n+1}{n}=\frac{1}{n}\frac{n+1}{2}=\frac{n+1}{2n}. \end{gather*}

Let us show by induction you tagged that for $n\ge 2$, $$\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)=\frac{n+1}{2n}.$$

For $n=2$, it is true since $1-\frac{1}{2^2}=\frac{3}{4}=\frac{2+1}{2\cdot 2}.$

Supposing that $\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)=\frac{n+1}{2n}$ gives you $$\begin{align}\prod_{k=2}^{n+1}\left(1-\frac{1}{k^2}\right)&=\frac{n+1}{2n}\cdot\left(1-\frac{1}{(n+1)^2}\right)\\&=\frac{n+1}{2n}\cdot\frac{(n+1)^2-1}{(n+1)^2}\\&=\frac{n(n+2)}{2n(n+1)}\\&=\frac{(n+1)+1}{2(n+1)}\end{align}$$