The Weierstrass product for reciprocal Gamma

Define $F(z):=z{e}^{\gamma z}{\prod }_{k\ge 1}(1+\frac{z}{k})e^{-z\mathrm{/}k}\mathrm{.}$ On a fixed compact set, after finitely many initial factors the logarithms of the remaining factors are $O(k^{-2})$ uniformly in $z$. Hence the product converges locally uniformly and defines an entire function. Its tail is zero-free, so its zeros are simple and occur exactly at $0,-1,-2,\dots$.

Let $x>0$. By [L3], $\mathrm{\Gamma }(x)$ is a positive real number. Taking reciprocals in [L1] is therefore valid at $x$, and rewriting the finite product gives $\frac{1}{\mathrm{\Gamma }(x)}={\lim }_{n\to \mathrm{\infty }}x{e}^{(H_n-\log n)x}{\prod }_{k=1}^n(1+\frac{x}{k})e^{-x\mathrm{/}k}=F(x),$ where [L2] supplies $H_n-\log n\to \gamma$.

On the right half-plane, both $F$ and $\mathrm{\Gamma }$ are holomorphic by [L4], so $F\mathrm{\Gamma }-1$ is holomorphic there. Step 2.1 makes it vanish on the positive real axis, which has accumulation points in that half-plane. Thus [L5] gives $F(z)\mathrm{\Gamma }(z)=1(\mathrm{Re}z>0)\mathrm{.}$

By [L4], the only poles of $\mathrm{\Gamma }$ are simple poles at the nonpositive integers. Step 1.1 gives $F$ a simple zero at each of those points, so every possible singularity of $F\mathrm{\Gamma }$ there is removable. The resulting entire function equals $1$ on the right half-plane by step 3.1, hence equals $1$ on all of $\mathbb{C}$ by [L5]. Therefore $F$ is the entire reciprocal of Gamma, which proves the displayed product formula everywhere.