Rouché's Theorem on $f_n(z)=1+\frac{1}{z}+\dots +\frac{1}{n!z^n}$ I want to prove that no matter how small we make $\epsilon>0$, for sufficiently great $n$, all the roots of $f_n(z)=1+\frac{1}{z}+\dots +\frac{1}{n!z^n}$ are situated inside the circle $|z|<\epsilon$.
This problem is solved here: Zeros of $f_n(z)=1+\frac{1}{z}+\frac{1}{2!z^2}+...+\frac{1}{n!z^n}$ are in $B_{\varepsilon}(0)$
But I was wondering if there was a way to directly apply the standard Rouché's theorem to $f_n(z)$ rather than to, though it looks like a really nice way of doing it, to $f_n(\frac{1}{z})$.
My thought: Applying Rouché's theorem directly to $f_n(z)$ doesn't seem as easy, because $f_n(z)$ converges to $e^{1/z}$ for all $z\neq 0$.  So, for a given $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that for $n>N$ we have $|f_n(z)-e^{1/z}|<\epsilon$. So, we need to have some circle $C$ such that the origin is not in in the interior or on the boundary of $C$..... now, do we have to play with our choice of $\epsilon$?  Is there a way to apply Rouché's theorem here?  Thank you!
 A: As requested in comments above, here's an approach to the problem using Hurwitz's theorem that's in the spirit of the question in the sense that it avoids using auxiliary functions. We'll use the below version of Hurwitz's theorem, which can be proved by the same methods as the version in Ahlfors. Going forward, for a connected region $\Omega$, let $N_f(\Omega)$ denote the number of zeroes with multiplicity of $f$ on $\Omega$.

Theorem (Hurwitz). Let $\Omega$ be a connected region with compact closure and $\Omega' \supset \overline{\Omega}$ be a connected region. Let $f_n$ be holomorphic on $\Omega'$ and $f_n \to f$ uniformly on compact subsets of $\Omega'$, with $f$ not uniformly zero. Then $\lim\limits_{n \to \infty} N_{f_n}(\Omega) = N_{f}(\Omega).$

With this in hand, let $\epsilon > 0$, $\Omega = S^2 - \overline{B(0, \epsilon/2)}, \Omega' = S^2 - \overline{B(0, \epsilon/4)}$, with $S^2$ the Riemann sphere where $\mathbb{C}$ is identified with $S^2 - \infty$ as usual. Denote $f(z) = \exp(1/z)$. Note also that each $f_n$ and $f$ is bounded in a neighborhood of infinity and $\lim_{|z|\to \infty} f_n(z) = \lim_{|z| \to \infty} f(z) = 1$. So by the removable singularity theorem, each $f_n$ and $f$ admits a holomorphic extension to $S^2 - 0$ with $f_n(\infty) = f(\infty) = 1$. By inspection $f_n(z) \to f(z)$ uniformly on $\Omega'$, and $f(z)$ has no zeroes on $\Omega$. By Hurwitz's theorem, $\lim\limits_{n \to \infty} N_{f_n}(\Omega) =N_f(\Omega) = 0$, so for all $n$ sufficiently large, $f_n$ is nonzero on the complement of $B(0, \epsilon)$ as desired.
