Hot to simplify $(1-\frac{1} {N} )(1-\frac{2} {N} )...(1-\frac{r-1} {N} )$ I'm trying to figure out the algebraic simplification from step $6$ to $7$ here 
Specifically the left side:
$(1-\frac{1} {N} )(1-\frac{2} {N} )...(1-\frac{r-1} {N} )$
To that:
$1-(\frac{1}{N}+\frac{2}{N} +... +\frac{r-1}{N}) +...$
I tried to multiply out but can't see how they got it and what are the terms after the last three dots. 
Thanks. 

In the article the birthmate problem is compared with birthday problem. In step $6$ it is obtained:
$$(1-\frac{1} {N} )(1-\frac{2} {N} )...(1-\frac{r-1} {N} )=\left(1-\frac1N\right)^n$$
Next it is stated: 

If we multiply out the left-hand side of Equation $(6)$ to terms of order $1/N$, and
  expand the right-hand side to two terms we get:
  $$1-(\frac{1}{N}+\frac{2}{N} +... +\frac{r-1}{N}) +\cdots=1-\frac1N+\cdots$$ 

Indeed, for the left side, consider simpler cases to see the pattern (up to the term $1/N$):
$$\begin{align}(1-\frac{1} {N} )(1-\frac{2} {N} )&=1-\frac1N-\frac2N+\frac{1\cdot 2}{N^2};\\
(1-\frac{1} {N} )(1-\frac{2} {N} )(1-\frac{3} {N} )&=\left(1-\frac1N-\frac2N+\frac{1\cdot 2}{N^2}\right)(1-\frac{3} {N} )=\\
&=1-\frac1N-\frac2N-\frac3N+\frac{1\cdot 2}{N^2}+\frac{1\cdot 3}{N^2}+\frac{2\cdot 3}{N^2}-\frac{1\cdot 2\cdot 3}{N^3};\end{align}$$
For the right side, the binomial theorem is used (up to two terms):
$$\left(1-\frac1N\right)^n=1-\frac{n}{1}\cdot \frac1N+\frac{n(n-1)}{2}\cdot \frac1{N^2}-\cdots$$
