Show that $| z_{k}|$ is increasing when $z_k$ is the zero of partial sum $P_{k}$ of $e^z$ with maximal absolute value.
This is the statement I'm having trouble proving. Can someone help me out here? I've tried applying Liouville Theorem but I'm stuck.
$P_k'(z)=P_{k-1}(z)$ so the zeroes of $P_k(z)$ are critical points of $P_{k+1}(z)$ and we can assume $k \ge 1$ as $P_0$ has no zeroes. Also notice that $P_k$ has only simple zeroes since $P_{k+1}(z)-P_k(z)=\frac{z^{k+1}}{(k+1)!}$ so the only possible multiple zero would be $0$ and that doesn't work.
Gauss Lucas Theorem implies that the zeroes of $P_k$ are strictly inside the convex hull of the zeroes of $P_{k+1}$ so they are of the form $\sum_{m=1}^{k+1} a_mz_m, a_m > 0 \sum {a_m}=1$ where $z_m$ are the zeroes of $P_{k+1}, m=1..k+1$.
This trivially implies that all the zeroes of $P_k$ are strictly less in absolute value than the largest absolute value zero of $P_{k+1}$ by the triangle inequality and the fact that $a_m >0, k+1 \ge 2$ so there are at least two distinct roots of $P_{k+1}$
Proof of Gauss Lucas for polynomials with simple roots - for non-simple roots theorem follows as here if $\zeta$ critical point is not a root, while it is obvious for critical points that are also roots - so double or higher such -
For any $P$ of degree $n \ge 2$ with simple roots $z_1,..z_n$, and $\zeta$ a critical point we have:
$0=\frac{P'}{P}(\zeta)=\sum \frac{1}{\zeta-z_m}=\sum \frac{1}{\bar \zeta-\bar z_m}$, hence
$\sum \frac{\zeta-z_m}{|\zeta-z_m|^2}=0$ which leads immediately to Gauss Lucas by separating the $\zeta$ term and dividing by $\sum \frac{1}{|\zeta-z_m|^2}$