Multiple integrals problem 
Show that!
  $$\int\limits_0^1
\int\limits_0^1
\int\limits_0^1
\int\limits_0^1
\int\limits_0^1
\int\limits_0^1
\frac{\,\mathrm du
\,\mathrm dv
\,\mathrm dw
\,\mathrm dx
\,\mathrm dy
\,\mathrm dz}{1-uvwxyz}=\frac{\pi^6}{945}$$

I have no idea about this problem. How to solve this multiple integrals? I have strucked on this problem, please anyone solve it for me.
 A: Notice that you can write
$$\frac{1}{1-uvwxyz}=\sum_{k=0}^{\infty} (uvwxyz)^k$$
since $0<u,v,w,x,y,z<1$.
So the integral is:
$$\sum_{k=0}^{\infty} \int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1(uvwxyz)^k\,du\,dv\,dw\,dx\,dy\,dz=\sum_{k=0}^{\infty} \frac{1}{(k+1)^6}=\zeta(6)$$
$$=\boxed{\dfrac{\pi^6}{945}}$$
A: $$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\mathrm{d}u\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z}{1-uvwxyz}\\
&=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{0}{\left(uvwxyz\right)}}{uvwxyz}\mathrm{d}u\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{1}{\left(vwxyz\right)}}{vwxyz}\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(wxyz\right)}}{wxyz}\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{3}{\left(xyz\right)}}{xyz}\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{4}{\left(yz\right)}}{yz}\mathrm{d}y\mathrm{d}z\\
&=\int_{0}^{1}\frac{\operatorname{Li}_{5}{\left(z\right)}}{z}\mathrm{d}z\\
&=\operatorname{Li}_{6}{\left(1\right)}\\
&=\frac{\pi^6}{945}\\
\end{align}$$
