Why $\prod_{i=1}^{k-1} \left( 1 - \frac{i}{N}\right) = 1 - \frac{ k \choose 2 }{N}$

Question

In this biology textbook, they show the following equation:

$$\prod_{i=1}^{k-1} \left( 1 - \frac{i}{N} \right) = 1 - \frac{ k \choose 2 }{N}$$

where both $N$ and $k$ are positive integers and $k < N$.

I fail to prove this equation correct. Can you help me?

Answer 1

What is true is that, for fixed $k$ as $N \to \infty$, $$ \prod_{i=1}^{k-1} \left(1 - \frac{i}{N}\right) = 1 - \frac{{k \choose 2}}{N} + O(1/N^2) $$

EDIT: Taking a few more terms,

$$ \eqalign{\prod_{i=1}^{k-1} \left(1 - \frac{i}{N}\right) &= 1 - \frac{{k \choose 2}}{N} + \dfrac{k(k-1)(k-2)(3k-1)}{24 N^2} - \frac{k^2 (k-1)^2 (k-2)(k-3)}{48 N^3}\cr & + \dfrac{k(k-1)(k-2)(k-3)(k-4)(15k^3-30k^2+5k+2)}{5760 N^4} + O(1/N^5)} $$

Answer 2

Bringing the generic factor to the common denominator, we have: $$ 1 - {i \over N} = {N-i \over N} $$ So, I am getting this: $$ \Pi_{i=2}^{k-1} {N-i \over N} = {N-2 \over N} \; {N-3 \over N} \; {N-4 \over N} \; \ldots \; {N-k+1 \over N} = { {(N-2)! \over (N-k)!} \over N^{k-2}} $$