I would like to construct a function that is holomorphic on $\mathbb{C}\setminus \mathbb{N}$ with a pole of order $n$ at each positive integer $n$. My idea was to take the following $f$: $$f(z)=\sum_{n\ge 1} \frac{(z-n)^{-n}}{n!}.$$ Is it clear that $f$ has a pole of order $n$ at each positive integer $n$. I would now like to show that it is holomorphic on $\mathbb{C}\setminus \mathbb{N}$. I believe it should be sufficient to show that the series representation of $f$ converges uniformly on every compact subset of $\mathbb{C}\setminus \mathbb{N}$. Indeed, for a fixed $R, \varepsilon>0$ let $$B(R,\varepsilon)=\left\{z\in\mathbb{C}: |z|\le R\right\} \cap \{z\in\mathbb{C}: |z-n|\ge \varepsilon \operatorname{for all} n\in\mathbb{N} \}\subset \mathbb{C}\setminus \mathbb{N},$$ where $B(R, \varepsilon)$ is compact.
For any $z\in B(R,\varepsilon)$, we have that $|z-n|\ge \varepsilon$, thus $|z-n|^{-n}\le \varepsilon^{-n}$ but $$\sum_{n\ge 1} \frac{\varepsilon^{-n}}{n!}$$ converges for any $\varepsilon>0$, thus by the M-test, we have uniform convergence and $f$ is holomorphic on $\mathbb{C}\setminus \mathbb{N}$.
My questions are:
(1) is the proof correct
(2) What are alternative ways of proving the desired statemnt?