Prove $\log|e^z-z|\leq |z|+1$

\begin{align*} |e^z-z|&\le|e^z|+|z|\\ &\le e^{|z|}+|z|\quad\textrm{from series expansion}\\ &\le e^{|z|+1}\quad\textrm{again from series expansion} \end{align*}

Use the definition: $e^z = \sum_{n=0}^\infty z^n/n!$

$$ \begin{align} \vert e^z - z \vert &= \vert 1+z^2/2 + z^3/6 + \cdots \vert \\ &\leq 1+\vert z \vert ^2/2 + \vert z^3 \vert/6 + \cdots\\ &= e^{\vert z \vert} - \vert z \vert \\ &\leq e^{\vert z \vert}\\ &< e^{\vert z \vert + 1} \end{align} $$