What is the range of the Gamma function I know from the product definition $\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}$ that it has no zeros. Is $0$ the only value omitted by $\Gamma$? I guess another way of looking at the question is that $\frac{1}{\Gamma(z)}$ is an entire function, so (by little Picard) omits at most one complex value; so the question is whether this function is surjective or whether it omits one complex value (if the latter, do we know which one)?
I have seen a similar question  for Riemann's zeta function answering that $\zeta$ is surjective, but I haven't been able to find anything for the Gamma function (my motivation is that I just learnt little Picard so I'm reviewing several basic functions from complex analysis, like all the trig functions, exponential, etc and trying to know the ranges of more of these functions).
 A: Note first that Picard theorem is valid for meromorphic functions in the plane with the statement that they can omit at most two values (finite or $\infty$); if the function is entire it obviously omits infinity, so one gets the usual statement about omitting at most one finite value.
In this case, as $1/\Gamma$ is entire hence omits infinity, so obviously $\Gamma$ omits zero and since $\Gamma$ is conjugate invariant it can omit at most another real value (if it would omit a complex non-real, it would omit its conjugate etc).
But the image of $(0, \infty)$ under it is $(1, \infty)$ and then looking on the negative axis we have that $\Gamma$ is negative on $(-2k-1,-2k)$ and positive on $(-2k-2, -2k-1)$ and going to the appropriate infinity at the ends; letting $a_{k}=\Gamma(x_k)<0$ the maximum of it on $(-2k-1,-2k)$ and $b_k=\Gamma(y_k)>0$ the minimum on $(-2k-2, -2k-1)$, one has that $b_{k+1} \le \Gamma(x_k-1)=a_k/(x_k-1), k \ge 1$ so since $|x_k-1| >3$  as $k \ge 1$ one has $b_{k+1} < |a_k|/3$
Similarly $|a_{k+1}| \le |\Gamma(y_k-1)|=b_k/|y_k-1| \le b_k/3$ which imply that $a_k, b_k \to 0, k \to \infty$ so $\Gamma$ cannot omit any other real value, hence its image is $\mathbb C \cup \infty - 0$
