Math Induction $N$ greater than or equal to $2$ $(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{n})$=$(\frac{1}{n})$ So I'm trying to make it equal.
so $n$ is equal or greater than 2.
When i substitute $2$ to n, $(1-\frac{1}{2})$ = $(\frac{1}{2})$ 
then
$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{k})$=$(\frac{1}{k})$
$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{k})+(1-\frac{1}{K+1})$=$(\frac{1}{k+1})$
I can't make it equal 
I got an answer of 
$\frac{1}{k}+\frac{k}{k+1}=\frac{1}{k+1}$ I don't know if N greater than or equal to 2 has something to do with it. The usual problems that I solve just prove n is equal to 1. I did the algebra and still can't make my LHS equal to $\frac{1}{k+1}$

As yashg mentions, you are adding and not multiplying. This might help:
$$\left(\prod_{k=2}^n\left(1-\frac1k\right)\right)\left(1-\frac1{n+1}\right)=\prod_{k=2}^{n+1}\left(1-\frac1k\right)$$
And
$$\begin{align}\left(\frac1n\right)\left(1-\frac1{n+1}\right)&=\frac1n-\frac1{n(n+1)}\\
&=\frac{n+1}{n(n+1)}-\frac1{n(n+1)}\\&
=\frac{n+1-1}{n(n+1)}\\
&=\frac{1}{n+1}\end{align}$$
