I want to understand the proof of
$$\prod_{k=2}^n \big(1- \frac{2}{k(k+1)} \big) = \frac{1}{3} \big(1+\frac{2}{n} \big), n \geq 2$$
via induction.
$$\text{Base case: } n=2 \Rightarrow\prod_{k=2}^2\big(1-\frac{2}{k(k+1)}\big) = 1-\frac{2}{2(2+1)} = \frac{2}{3} = \frac{1}{3} \big(1+\frac{2}{2}\big)$$
$$\text{Hypothesis: The proposition holds for an } n \in \mathbb{N} \text{ with } n \ge 2$$
$$\text{Induction step: } \prod_{k=2}^{n+1} \big(1- \frac{2}{k(k+1)} \big) = \big(1- \frac{2}{(n+1)(n+2)} \big) \cdot \prod_{k=2}^n \big( 1- \frac{2}{k(k+1)} \big)$$
$$= \text{IH: } \big( 1- \frac{2}{(n+1)(n+2)} \big) \cdot \frac{1}{3} \big(1+ \frac{2}{n}\big) $$
$$= \frac{(n+1)(n+2)-2}{(n+1)(n+2)} \cdot \frac{1}{3} \cdot \frac{n+2}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
$$=\frac{1}{3} \frac{n^2+3n+2-2}{n(n+1)} = \frac{1}{3} \cdot \frac{n+3}{n+1 }= \frac{1}{3}\cdot \big(\frac{n+1}{n+1} +\frac{2}{n+1}\big)$$
$$=\frac{1}{3} \big(1+\frac{2}{n+1} \big)$$
What I don't understand are the two steps:
What's been done to remove the $1-$ and $1+$?:
$$= \frac{(n+1)(n+2)-2}{(n+1)(n+2)} \cdot \frac{1}{3} \cdot \frac{n+2}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
How do we get the $\frac{1}{3}\cdot \big(\frac{n+1}{n+1} +\frac{2}{n+1}\big)$ here?:
$$=\frac{1}{3} \frac{n^2+3n+2-2}{n(n+1)} = \frac{1}{3} \cdot \frac{n+3}{n+1 }= \frac{1}{3}\cdot \big(\frac{n+1}{n+1} +\frac{2}{n+1}\big)$$
In the first step we are using the induction hypotesis that is
$$\prod_{k=2}^n \left(1- \frac{2}{k(k+1)} \right) = \frac{1}{3} \left(1+\frac{2}{n} \right)$$
the second one is
$$= \frac{(n+1)(n+2)-2}{(n+1)\color{red}{(n+2)}} \cdot \frac{1}{3} \cdot \frac{\color{red}{n+2}}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
and finally
$$=\frac{1}{3} \frac{n^2+3n+2-2}{n(n+1)} =\frac{1}{3} \frac{n(n+3)}{n(n+1)}= \frac{1}{3} \cdot \frac{n+3}{n+1 }=$$$$= \frac{1}{3} \cdot \frac{n+1+2}{n+1 }= \frac{1}{3}\cdot \left(\frac{n+1}{n+1} +\frac{2}{n+1}\right)$$