Consider the complex-valued family of functions $$ f_n(z) = \sum_{k=0}^n \frac{1}{k!}z^k. $$
Is it possible to use Rouché's Theorem to prove that for any $R \in \mathbb R$ there exists some $n$ such that all roots of $f_n$ have modulus strictly greater than $R$?
The solutions that I've seen for this exploit the uniform convergence of the Taylor series for $e^z$, but I'm curious if it's possible to use Rouché's Theorem to show that there are no roots inside the circle of radius $R$ for sufficiently large $n$.