Poles of the gamma function

Remarks

Iterating the functional equation gives $\Gamma(z) = \dfrac{\Gamma(z + n + 1)}{z (z + 1) \cdots (z + n)}$ near $z = -n$. The numerator is holomorphic at $-n$ with value $\Gamma(1) = 1$, and only the factor $z + n$ of the denominator vanishes there, to first order. So the pole is simple, and the residue is $1/\big((-n)(-n + 1) \cdots (-1)\big) = (-1)^n/n!$. Away from these points $\Gamma$ is holomorphic, see holomorphy of the gamma function. For example, near $z = 0$ one has $\Gamma(z) = \frac{1}{z} - \gamma + O(z)$, where $\gamma$ is the Euler–Mascheroni constant.

Going deeper

For a student

The residues explain contour integral formulas such as the Mellin–Barnes integrals, where closing the contour to the left picks up the poles of $\Gamma$ and produces power series. For instance, summing the residues of $\Gamma(s) x^{-s}$ at $s = -n$ gives $\sum_n (-1)^n x^n / n! = e^{-x}$. The fact that the poles are simple and $\Gamma$ has no zeros makes $1/\Gamma$ an entire function with simple zeros.