Proving an inequality involving the logarithm In Ahlfors' complex analysis text he claims that:

If $|u| < 1$ we have by power-series development
   $$ \log \left\lvert E_h(u) \right\rvert \leq \frac{1}{h+1} |u|^{h+1}+\frac{1}{h+2}|u|^{h+2}+\dots$$

where $$E_h(u)=(1-u)e^{u+\frac{1}{2}u^2+ \dots\frac{1}{h}u^h} $$
I can't see why is this inequality true. I want to use the power-series of the logarithm, but the absolute value gets in the way.
P.S. $u$ is complex and $h$ is a positive integer.

The logarithm of the absolute value of a complex number $\neq 0$ is just the real part of the logarithm, whichever branch of the logarithm is chosen.
For $\lvert u\rvert < 1$, we can choose the principal branch, and have
$$\begin{align}
\log E_h(u) &= \left(u + \frac12u^2 + \dotsb \frac{1}{h}u^h\right) + \log (1-u)\\
&= \sum_{k=1}^h \frac{u^k}{k} - \sum_{k=1}^\infty \frac{u^k}{k}\\
&= -\sum_{k=h+1}^\infty \frac{u^k}{k}.
\end{align}$$
Now estimate the real part by the absolute value to get
$$\log \lvert E_h(u)\rvert \leqslant \sum_{k=h+1}^\infty \frac{\lvert u\rvert^k}{k}.$$
