Complex Conjugate of Gamma Function

Theorem
Let $\Gamma$ denote the gamma function.

Then:
 * $\forall z \notin -\N_0: \Gamma \left({\overline{z}}\right) = \overline {\Gamma \left({z}\right)}$

where $N_0 = \N \cup \left\{{0}\right\}$.

Proof
This is immediate from, say, the Euler form of $\Gamma$ and the fact that complex conjugation preserves products and sums.