Seeking possibility of more elementary means of evaluating an improper integral. It can be shown that $\int_0^\infty -\log{(1-e^{-x})}=\zeta(2)$ by expanding out the integral as $\log(1-z)$, exchanging summation and integration, then summing up the integrals. I am wondering if there are means to prove this that do not require such advanced tools.
This integral also represents the area bound by the axes and the curve $e^{-x}+e^{-y}=1$, in case that fact is of any assistance.

The two approaches that come to mind both use the power series for $-\log(1-x)$.
First Approach: Use the power series for $-\log(1-x)$
$$
\begin{align}
\int_0^\infty-\log\left(1-e^{-x}\right)\mathrm{d}x
&=\int_0^\infty\sum_{k=1}^\infty\frac{e^{-kx}}k\,\mathrm{d}u\\
&=\sum_{k=1}^\infty\frac1k\int_0^\infty e^{-kx}\,\mathrm{d}u\\
&=\sum_{k=1}^\infty\frac1{k^2}\\[6pt]
&=\zeta(2)
\end{align}
$$
Second Approach: Substitute $u=e^{-x}$
$$
\begin{align}
\int_0^\infty-\log\left(1-e^{-x}\right)\mathrm{d}x
&=\int_0^1\frac{-\log\left(1-u\right)}u\,\mathrm{d}u\\
&=\int_0^1\sum_{k=1}^\infty\frac{u^{k-1}}k\,\mathrm{d}u\\
&=\sum_{k=1}^\infty\frac1k\int_0^1u^{k-1}\,\mathrm{d}u\\
&=\sum_{k=1}^\infty\frac1{k^2}\\[6pt]
&=\zeta(2)
\end{align}
$$
Both seem pretty elementary.
