sequences and series - Simpler derivation of $sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$

I know that the equality $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ can be proved in numerous ways by using the Fourier series. However, is there a way to derive it using more fundamental tools? I've tried:
$$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^1 x^{n-1} dx \int_0^1 y^{n-1} dy = \int_0^1 dx\int_0^1 dy \frac{1}{1-xy}$$
and by changing variables I was able to write it in several other forms: $$= -\int_0^1\frac{\ln(1-x)}{x} dx = \int_0^\infty \frac{\ln (1+t)}{t(1+t)} dt = \int_0^\infty \frac{u }{e^u -1}du$$
but that's as far as I could get.

I specifically don't want to use Fourier series. More fundamental complex analysis, like contour integration, is fine.

Answer

Yes, this can be done. Actually, that's what Tom Apostol did in an article he published in 1983; you can read the proof here (it's the first proof). He used a change of variable ($(x,y)\mapsto(x+y,x-y)$) in order to compute the integral $$\iint_{[0,1]\times[0,1]}\frac1{1-xy}\,\mathrm dx\,\mathrm dy.$$