Is the following gamma function integral correct? $$\int_{0}^{\infty}\frac{x^{v-1}}{e^{-\beta\mu}{e^x}-1}=\Gamma(v)\left(e^{\beta\mu}+\frac{e^{2\beta\mu}}{2^v}+\frac{e^{3\beta\mu}}{3^v}+\dotsb\right)$$

This integral is given in my statistical physics assignment and I feel that the $\mu$ in the integrand should have the opposite sign (as my results would make more sense) but I don't know how to prove this and I can't find the formula anywhere. I would love to find a source for this formula.

There's no sign error. You just need to identify a hidden convergent geometric series:$$\int_0^\infty\frac{x^{v-1}e^{\beta\mu}e^{-x}dx}{1-e^{\beta\mu}e^{-x}}=\sum_{n\ge1}e^{n\beta\mu}\int_0^\infty x^{v-1}e^{-nx}dx=\Gamma(v)\sum_n\frac{e^{n\beta\mu}}{n^v}.$$Incidentally, this can be rewritten with the polylogarithm as $\Gamma(v)\operatorname{Li}_v(e^{\beta\mu})$; convergence requires $\Re v>1,\,\Im\beta\mu<0$.
