Prove: For every $\delta \gt 0$, there is a natural number $n$, such that all the zero points of $1 + \frac 1 z + \frac 1 {2!z^2} + \cdots + \frac 1{n!z^n}$ satisfy $\lvert z \rvert \lt \delta$.
Then the answer goes:
If $\delta$ is small enough, $$e^z = 1 + z + {z^2 \over 2!}+ \cdots + {z^n \over n!}+ \cdots$$ is uniformly convergent on $\lvert z \rvert \le \delta^{-1}$.
So is $$e^{\frac 1 z} =1 + \frac 1 z + \frac 1 {2!z^2} + \cdots + \frac 1{n!z^n} + \cdots$$ on $\lvert z \rvert \ge \delta$.
By Cauchy's convergence test, there exists one $N = N(\delta)$, such that when $n \gt N$, $\lvert{\sum_{k = n+1}^\infty {1 \over {k!z^k}} \rvert}\lt \frac m 2$ $(\lvert z \rvert \ge \delta)$. Then $$\inf_{\lvert z \rvert \ge \delta} \lvert e^{1 \over z}\rvert = \inf_{\lvert z \rvert \le \delta^{-1}} \lvert e^z\rvert \equiv m.$$(I don't know how this equation is drawn.I don't know much about infimum ...)
So, $$\lvert 1 + \frac 1 z + \frac 1 {2!z^2} + \cdots + \frac 1{n!z^n} \rvert \ge \lvert e^{1 \over z}\rvert - \lvert\sum_{k = n+1}^\infty {1 \over {k!z^k}} \rvert \gt m - \frac m 2 = \frac m 2 \gt 0(\lvert z \rvert \ge \delta).$$
Qed.