Multiple integral related to zeta function I am trying to calculate the following integral
$$\int_V \frac{d^d\vec{r}}{e^{x_1+...+x_d}-1},$$
where $V=[0,\infty)^d$ and $\vec{r}=(x_1,...,x_d)$. I know that the result should be related to the Riemann zeta function, but I do not see how to do it quickly and elementary (i.e. without the knowledge of all possible relations for zeta function). Any sugestion or hint ?

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\begin{align}
&\left.\int_{\large\left[0,\infty\right)^{d}}
{\dd^{d}\vec{r} \over \expo{x_{1} + \cdots + x_{d}} - 1}\,
\right\vert_{\ d\ \geq\ 2}
\\[5mm] = &\
\int_{\large\left[0,\infty\right)^{d}}
\pars{\int_{0}^{\infty}
\braces{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\exp\pars{\bracks{z - x_{1} - \cdots - x_{d}}s}\,{\dd s \over 2\pi\ic}}
{\dd z \over \expo{z} - 1}}\dd^{d}\vec{r}
\\[5mm] = &\
\int_{0}^{\infty}{1 \over \expo{z} - 1}
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\expo{zs}
\pars{\int_{0}^{\infty}\expo{-xs}\,\dd x}^{d}{\dd s \over 2\pi\ic}\,\dd z =
\int_{0}^{\infty}{1 \over \expo{z} - 1}
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{zs} \over s^{d}}
{\dd s \over 2\pi\ic}\,\dd z
\\[5mm] = &\
\int_{0}^{\infty}{1 \over \expo{z} - 1}{z^{d - 1} \over \pars{d - 1}!}\,\dd z =
{1  \over \pars{d - 1}!}\sum_{n = 0}^{\infty}
\int_{0}^{\infty}z^{d - 1}\expo{-\pars{n + 1}z}\,\dd z
\\[5mm] = &\
{1  \over \pars{d - 1}!}
\sum_{n = 0}^{\infty}{\pars{d - 1}! \over \pars{n + 1}^{d}} =
\bbx{\zeta\pars{d}}
\end{align}
