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Proving a certain identity involving the exponential function
Prove the following identity involving the exponential function
$$\frac{1}{1-e^{ -x}}=\sum_{k=1}^\infty \frac{x^k}{k!} $$
where $x\in \mathbb{R} $
Hint: Consider the integral $\int_0^x \frac{1}{1-e^{ -u}}du$
I have some doubts regarding the hint.
\begin{align}
\int_0^x \frac{1}{1-e^{ -u}}du & = \ln \left| \frac{1-e^{ -x}}{1-e^{ -0}} \right|
= \ln \left| 1-e^{ -x}\right|
& \leq \ln \left| e^{ -x}\right| + \ln \left| 1-e^{ -0}\right| \\
& \leq \ln \left| e^{ -x}\right| + \ln 2
\end{align}
Do I need to justify that the value of the last inequality is $x$?
Could someone give me a hand with the proof?
A:
The hint is quite unclear. You should notice that
$$
\int_0^x\frac{u^k}{k!}du=\frac{x^{k+1}}{k+1}.
$$
All in all:
$$
\sum_{k=0}^\infty\frac{x
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