On the page 311 in the book of Flajolet appears that $(1-a)^n=e^{-an}\exp(O(na^2))$. I am trying to understand this formula and, in particular, the term $\exp(O(na^2))$.
I've tried the following. From binomial expansion we have \begin{align*} (1-a)^n&=1+\binom{n}{1}(-a)+\binom{n}{2}(-a)^2+\binom{n}{3}(-a)^3+\ldots\\ &=1+\frac{(-a)n}{1!}+\frac{(-a)^2n^2}{2!}+\frac{(-a)^2(-n)}{2!}+\frac{(-a)^3n^3}{3!}+\frac{(-a^3)(-3n^2+2n)}{3!}+\ldots \end{align*}
The last line can be written as \begin{align*} \left(1+\frac{(-a)n}{1!}+\frac{(-a)^2n^2}{2!}+\frac{(-a)^3n^3}{3!}+\ldots\right)\cdot\left(1-\frac{a^2n}{2!}+\frac{-a^3(2n)}{3!}+(-a)^4n^2\left(\frac{11}{4!}-\frac{1}{3}\right)+\ldots\right), \end{align*} so that inside the first parenthesis we get $e^{-na}$, which means that the infinite sum inside the second parenthesis should produce $\exp(O(na^2))$. Could you please explain this? How do we obtain this $O$-expression?
Both the dominant term $e^{-an}$ and the error term are exponentials, so naturally you take logarithms. That is, you want to show
$$\ln (1 - a)^n = n \ln (1 - a) = -na + O(na^2).$$
But this follows from e.g. Taylor's theorem applied to the Taylor series of $\ln (1 - a) = a - \frac{a^2}{2} \pm \dots $, at least for $a \in (-1, 1)$, which I think is all Flajolet needs. We do not need to assume that $a$ is fixed as $n \to \infty$ as long as $a$ is bounded away from $-1$ and $1$; this approximation can usefully be applied if, for example, $a \sim \frac{x}{n}$ with $x$ fixed, where it gives $e^{-x} \exp O(\frac{1}{n})) = e^{-x} ( 1 + O(1/n) )$.