Let $$ h(z):=\sum\limits_{k=0}^\infty \frac{i^{k+1}z^k}{(k+1)!} $$
$h$ is entire, so it is continuous. I saw the following argument [somehow] derived from continuity of $h(z)$:
$\dots$ In particular, $h$ is continuous at $0$, so there exists $\delta>0$ such that $|z-0|<\delta$ implies $|h(z)|<2|h(0)|\dots$.
Here's how I would say:
In particular, $h$ is continuous at $0$, so there exists $\delta>0$ such that $|z-0|<\delta$ implies $|h(z)-h(0)|<\varepsilon$.
So that $h(0)-\varepsilon<h(z)<h(0)+\varepsilon$. Where's the $2$ before $|h(0)|$ coming from?
$|h(0)|=1 >0$
Hence by continuity, let $\epsilon = |h(0)|$, $\exists \delta >0$ such that $|z-0| < \delta$ implies $$|h(z)-h(0)| < |h(0)|$$
Hence $|h(z)| = |h(z)-h(0)+h(0)| \leq |h(z)-h(0)| + |h(0)| < 2|h(0)|$