Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{(k+1)^2}{k(k+2)}$

First observe that the product is a telescopic product: $$\prod_{k = 1}^n \frac{(k+1)^2}{k(k+2)} = \frac{2^2}{3} \cdot \frac{3^2}{2\cdot 4} \cdot \frac{4^2}{3 \cdot 5} \cdot ... \cdot \frac{n^2}{(n-1)(n+1)} \cdot \frac{(n+1)^2}{n(n+2)} = \frac{2(n+1)}{n+2}. $$ In case of need use induction to prove it.

Now it is easy: $$ \lim_{n \to \infty} \frac{2(n+1)}{n+2} = 2.$$

Using $$\prod_{k=1}^{n} (k+1) = \prod_{k=2}^{n+1} k = (n+1)!$$ then, with similar products, \begin{align} L &= \lim_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} \\ &= \lim_{n \to \infty} \frac{2 \, ((n+1)!)^2}{n! \, (n+2)!} \\ &= \lim_{n \to \infty} \frac{2 (n+1)}{n+2} = 2 \end{align}