Everything Cancels Off But One, Right?

Using $\small{\color{#3D99F6}{x^2-y^2=(x+y)(x-y)}}$ $\large \mathbf{P}=\displaystyle\prod_{n=2}^\infty \dfrac{n\cdot n}{(n-1)(n+1)}$ $=\left(\displaystyle\prod_{n=2}^m \dfrac{n}{n-1}\right)\times \left(\displaystyle\prod_{n=2}^m \dfrac{n}{n+1}\right)$ $=(\dfrac{\cancel 2}{1}\times \dfrac{\cancel 3}{\cancel 2}\times \dfrac{\cancel 4}{\cancel 3}\times \cdots \dfrac{m}{\cancel{m-1}})\times (\dfrac{2}{\cancel 3}\times \dfrac{\cancel 3}{\cancel 4}\times \dfrac{\cancel 4}{\cancel 5}\times \cdots \dfrac{\cancel m}{m+1})$ $=\dfrac{2m}{m+1}$ When $m$ $\rightarrow \infty,\color{forestgreen}{\mathbf P~\text{approaches}~2}$

Hence, $\Large\color{#302B94}{\mathbf P=2}$

Taking log both side apply property and express it as difference of two consecutive term the replace upper limit by term wise higher term and lower term with lower limit and at last take antilog to get answer