Prove the inequalities $|e^{x}-1|\leq e^{|x|}-1\leq |x|e^{|x|}$ Prove that $|e^{x}-1|\leq e^{|x|}-1\leq |x|e^{|x|}$ for all $x\in \mathbb{C}$. I did this by Maclaurin series of $e^x$,
$$|e^{x}-1|\leq |x|+\frac{|x|^2}{2!}+\frac{|x|^3}{3!}+\mathcal{O}(|x|^{4})=e^{|x|}-1 \\
\leq |x|^2+\frac{|x|^3}{2!}+\frac{|x|^4}{3!}+\mathcal{O}(|x|^{5})=|x|e^{|x|}.$$
My teacher disliked the last inequality, and I don't know why. Is there a way to fix this problem?

Well, you could use some steps in between:
\begin{align*}
e^{| x| } - 1 = \sum_{n=1}^\infty \frac{|x|^n}{n!} = | x | \sum_{n=1}^\infty \frac{|x|^{n-1}}{(n-1)!} \frac{1}{n!} \le |x| \sum_{n=1}^\infty \frac{|x|^{n-1}}{(n-1)!} = |x| e^{|x|}
\end{align*}
as $\frac{1}{n!} \le 1$ for all $n \ge 1$.
Note that after the inequality we have start with a  factor $|x|$ instead of $|x|^2$ as the Maclaurin series of $\exp(|x|)$ start with a $1$ and not with $|x|$.
