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. 
 A: In the link you give, Frederick Mosteller writes: 

If we multiply out the left-hand side of Equation (6) to terms of order $1/N$ [...]

So let us do that. Here at do it step by step to show the process, and write "..." for terms that are at least quadratic in $1/N$:
$$\begin{align*}
\left(1-\frac{1}N\right)\left(1-\frac{2}{N}\right)\cdots &\left(1-\frac{r-1}{N}\right)
= {\color{red}1}\cdot\left(1-\frac{2}{N}\right)\cdots \left(1-\frac{r-1}{N}\right)-{\color{red}{\frac{1}{N}} }\left(1-\frac{2}{N}\right)\cdots \left(1-\frac{r-1}{N}\right)\\
&= \left(1-\frac{2}{N}\right)\left(1-\frac{3}{N}\right)\cdots \left(1-\frac{r-1}{N}\right)-\frac{1}{N} + \dots\\
&= \left(1-\frac{3}{N}\right)\cdots\left(1-\frac{r-1}{N}\right)
- \frac{2}{N} \left(1-\frac{3}{n}\right)\cdots\left(1-\frac{r-1}{N}\right)-\frac{1}{N} + \dots\\
&= \left(1-\frac{3}{N}\right)\cdots\left(1-\frac{r-1}{N}\right)
- \frac{2}{N} + \dots -\frac{1}{N} + \dots\\
&= \left(1-\frac{3}{N}\right)\cdots\left(1-\frac{r-1}{N}\right)
- \frac{2}{N} -\frac{1}{N} + \dots\\
& \vdots\\
&= \left(1-\frac{r-1}{N}\right) - \frac{r-1}{N} - \cdots
- \frac{2}{N} -\frac{1}{N} + \dots\\
&= 1 - \left(\frac{1}{N}+\frac{2}{N}+\cdots+\frac{r-1}{N}\right) - \dots
\end{align*}$$
as claimed.
Basically, what it is doing is expanding the $(r-1)$-term product (there are $2^{r-1}$ terms), and only showing the term $1=\underbrace{1\cdot1\cdots 1}_{r-1\text{ times}}$ and the $r-1$ terms of the form $-\frac{k}{N}$ (obtained by multiplying $r-2$ factors being $1$, and one being $-\frac{k}{N}$).
A: 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):
$$\left(1-\frac1N\right)^n=1-\frac{n}{1}\cdot \frac1N+\frac{n(n-1)}{2}\cdot \frac1{N^2}-\cdots$$
