<s>[INST] 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. 
 [/INST] 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):
$$\left(1-\frac1N\right)^n=1-\frac{n}{1}\cdot \frac1N+\frac{n(n-1)}{2}\cdot \frac1{N^2}-\cdots$$
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