Integrate twice

We can write $\frac{1}{1-xy} = 1+(xy)+(xy)^{2}+(xy)^{3}+...$ Since $\int_{0}^{1} \int_{0}^{1}(xy)^{n} dx\space dy \space = (\int_{0}^{1}x^{n} dx)^{2}\ = (\frac{1}{n+1})^{2}$ we have $\int_{0}^{1} \int_{0}^{1}\frac{1}{1-xy} dx\space dy = \sum_{n=0}^{\infty}(\frac{1}{n+1})^{2} = \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ .