The following is taken from page 8 of Alon and Spencer's The Probabilistic Method.
$ \prod_{i = 0}^{n-1} \frac{v - 2i}{v-i} \sim e^{-n^2/2v} $ as long as $v \gg n^{3/2}$, estimating $\frac{v-2i}{v-i} = 1 - \frac{i}{v} + O\left(\frac{i^2}{v^2}\right) = e^{-i/v + O(i^2/v^2)}$
I'm able to derive the asymptotics expression on the first line given the equality on the second line, and it is in this derivation that the condition $v \gg n^{3/2}$ is used. What I do not understand:
Why is $\frac{v-2i}{v-i} =1 - \frac{i}{v} + O\left(\frac{i^2}{v^2}\right)?$ At a gut level, I see that $i$ is quite small compared to $v$, so the ratio should be a little less than $1$, but I think I proper derivation would be instructive to me.
Why is $1 - \frac{i}{v} + O\left(\frac{i^2}{v^2}\right) = e^{-i/v + O(i^2/v^2)}?$ I am familiar with the fact that $1 - x \sim e^{-x}$ for small, positive $x$, but what justifies taking $O(i^2/v^2)$ directly into the exponent?