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Suppose we have the function $\Gamma$ defined for all $z \in \mathbb{C}$ with $\mathrm{Re}~ z > 0$ by the standard integral formula $$\Gamma(z) = \int_{0}^{\infty} e^{-t} t^{z - 1} dt$$ (perhaps extended to $\mathbb{C}$ except $0,-1,-2,\ldots$ via recurrent relation), and that we aim to prove the identity $$\Gamma(z) = \lim_{n \to \infty} \frac{n^z n!}{z (z+1) \ldots (z + n)}.$$ In the real case, there is a relatively simple proof of this identity utilising log-convexity of $\Gamma$ (demonstrated, e.g., in Artin's Gamma Function).

Is there any simple way how to use the identity for real-variable $\Gamma$ (or the log-convexity argument) to obtain the identity in the complex case? It is trivial if one knows that both sides of the identity are analytic functions – but this seems to me as relatively non-trivial compared to the rest of the stuff.

My main motivation for asking this question is that I have found multiple sources, which effectively state the proof for $\Gamma$ over $\mathbb{R}$ and then claim that they have proved the property for $\Gamma$ over $\mathbb{C}$. I am therefore curious if there is some simple argument that can be used in the transtion from $\mathbb{R}$ to $\mathbb{C}$.

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    $\begingroup$ Does the answer here adequately address your question? $\endgroup$ – Eric Towers Dec 7 '19 at 16:12
  • $\begingroup$ @EricTowers Not really, I think. If I understand that answer correctly, this is just what I mention in the penultimate paragraph: if you know that both sides are analytic functions, the task is easy... $\endgroup$ – Jára Cimrman Dec 7 '19 at 17:24
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    $\begingroup$ Since the logarithm behaves differently on the complex plane I feel you should end needing analyticity anyway but just my intuition so not sure $\endgroup$ – Daniel D. Dec 8 '19 at 19:20

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