Prove for every $R>0$ there's an integer $n_0>0$ such that if $n\geq n_0$ then $f_n$ has no zeroes. Using Hurwitz Theorem. Let
$$f_n(z)=\sum_{k=0}^n \frac{z^k}{k!}$$
and let
$$f(z)=\sum_{k=0}^\infty \frac{z^k}{k!}=e^z.$$ As a polynomial $f_n$ has $n$ roots in $\Bbb{C}$. Prove for every $R>0$ there's an integer $n_0>0$ such that if $n\geq n_0$ then $f_n$ has no zeroes in $B_R(0)$. So I want to appeal to Hurwitz's theorem that States if $f_n \rightarrow f$ uniformly on every compact subset of $\Bbb{C}$ which the $f_n$ do converge uniformly to $e^z$ on every compact subset, then if $f$ has a zero at $z_0$ of order $m$ so do the $f_n$. Then since $f(z)$ has no zeroes, can I conclude the $f_n$'s have no zeroes? Or do I need to find such an $n_0$.

Fix $R>0.$ Since $f_n$ tends uniformly to $e^z$  on the disc $|z|\le R,$ there exists $n_0$ such that for $$|f_n(z)-e^z|< e^{-2R},\ n\ge n_0,\ |z|\le R$$
Then
$$|f_n(z)|\ge |e^z|- |f_n(z)-e^z|\\ \ge e^{-R}-e^{-2R}>0,\ n\ge n_0,\ |z|\le R$$
Remark No analytic function theory is needed. Almost uniform convergence to a function which does not vanish, suffices.
