I am reading a paper in which a proof say that for $n \geq 1$ and $|z| \leq A$ (where $z \in \mathbb{C}$) the following function is bounded as follows:
$$ \frac{\Gamma(n+z)}{\Gamma(n+1)}= \frac{n^z}{n+z} \text{exp} \left(z \cdot O\left( \frac{1}{n} \right)\right) \text{exp} \left( \sum_{k=n+1}^{\infty} \sum_{m=2}^\infty \frac{(-1)^m z^m}{m \cdot k^m}\right) $$
$$= \frac{n^z}{n+z} \text{exp}\left(O_A\left(\frac{1}{n}\right)\right) \left(1+O_A\left(\frac{1}{n}\right)\right) = n^{z-1} \left(1+O_A\left(\frac{1}{n}\right)\right) $$
The way to get to that first equality is fairly straightforard. And my understanding is that the notation $f(z)=O_A \left(g(z)\right)$ means that $|f(z)|\leq c |g(z)|$ for $z$ large enough and where $c$ depends of the parameter $A$. With that in mind, passing from $ \text{exp} \left(z \cdot O\left( \frac{1}{n} \right)\right)$ to $\text{exp}\left(O_A\left(\frac{1}{n}\right)\right)$ is trivial.
But I do not see how the other exponential simply turns into $\left(1+O_A\left(\frac{1}{n}\right)\right)$ or why that bound would depend of $A$, or how the last equality emerges.
Any help clearing up those last to equalities? Thanks in advance!