It can be shown that $$\int_0^\infty -\ln{(1-e^{-x})}dx=\zeta(2)$$ by expanding out the integral as $\ln(1-z)$, 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.
Here is an elementary procedure
\begin{align} I=&\int_0^\infty -\ln{(1-e^{-x})}\ dx\\ =& \ \frac12\int_0^\infty -\ln{{(1-e^{-2x})}} \overset{2x\to x} {dx}+ \frac12 \int_0^\infty\ln{ \frac{1+e^{-x}}{1-e^{-x}} } \overset{x\to 2x}{dx}\\ =& \ \frac14I+\int_0^\infty\ln\coth x \ dx =\frac43\int_0^\infty\ln\coth x\ dx\\ =& \ \frac43 \int_0^\infty \int_0^1\frac{t }{\cosh^2 x-t^2}dt \ dx =\frac43 \int_0^1 \frac{\sin^{-1}t}{\sqrt{1-t^2}}dt =\frac{\pi^2}6 \end{align}