I have to show that show that $\sum_{k = 1}^{\infty} \frac{1}{n! z^n}$ is analytic on $\mathbb{C}\{0\}$ and calculate its integral around the unit circle.
My attempt is to try and use the analytic convergence theorem but I am not sure how to prove uniform convergence first which is what I think I need to continue using the theorem. Any hints on how to approach the problem are appreciated.
On
Abridged solution. If $|z| > a$ then $\left| \dfrac{1}{n! z^n} \right| \leq \dfrac{1}{n! a^n}$ and the latter series converges whatever the value of $a > 0$ may be, hence the former series converges uniformly on each open set $\{|z| > a\}$ and consequently, the former series is analytic on $\mathbf{C}^*.$ Q.E.D.
Note that$$\sum_{n=1}^\infty\frac1{n!z^n}=-1+\sum_{n=0}^\infty\frac{\left(\frac1z\right)^n}{n!}=e^{\frac1z}-1,$$which is analytic on $\mathbb{C}\setminus\{0\}$. On the other hand, it is clear that the residue of your functions at $0$ is $1$. Therefore, the integral that you mentioned is $2\pi i$, by the residue theorem.