Prove this inequality for $|z| \leq 1$ I am supposed to prove that 
$$\frac{|e^z-1|}{e-1} \leq |z|$$ for $|z| \leq 1$. My guess is that I have to show that the LHS $\leq 1$ and then apply Schwarz's Lemma. But I am not able to prove that!
 A: Try $\mathrm e^z-1=z\int\limits_0^1\mathrm e^{tz}\,\mathrm dt$. Hence $|\mathrm e^z-1|\leqslant|z|\int\limits_0^1|\mathrm e^{tz}|\,\mathrm dt$. Since $|\mathrm e^{tz}|=\mathrm e^{t\Re z}\leqslant\mathrm e^{t|z|}\leqslant\mathrm e^t$, the integral is at most $\int\limits_0^1\mathrm e^{t}\,\mathrm dt=\mathrm e-1$, hence $|\mathrm e^z-1|\leqslant|z|\cdot(\mathrm e-1)$, as desired.
This method proves more generally that, for every $z$, $|\mathrm e^z-1|\leqslant|z|\cdot(\mathrm e^{|z|}-1)$.
A: Define $f(z) = \frac{e^z - 1}{e - 1}$. This is a holomorphic function, and it satisfies $f(0) = 0$. Thus, by Schwarz's Lemma, $|f(z)| \leq |z|$ for all $|z| < 1$.
Now all that remains is extending this result to $|z| \leq 1$. This is a simple technical result: The function $g(z) = |f(z)| - |z|$ is a continuous, real, non-positive* function defined in the entire complex plane. By continuity, it cannot attain a positive value at a point on the unit circle; otherwise, it would have to be positive at some neighborhood of that point, which includes points inside the disk.
*edit: I meant non-positive when $|z|<1$ as we've shown.
A: We can write 
\begin{align}
|e^z-1|&=\left|\sum_{n\geq 1}\frac{z^n}{n!}\right|\\
&\leq \sum_{n\geq 1}\frac{|z|^n}{n!}\\
&\leq |z|\sum_{n\geq 1}\frac 1{n!}\\
&=|z|(e-1)
\end{align}
