Found this identity by accident I found the following identity when trying to prove something else:
$$\sum\limits_{n=1}^\infty\frac{(n-1)!x}{\prod\limits_{k=1}^n (k+x)}\equiv 1,~ \forall x>0$$
I'm sure there's a name for it (or for a more general identity for which this is a special case). Does anyone know the name?
 A: There may be a more specific name for this formulae, but it follows almost immediately from Gauss' Formulae:
$$_2F_1(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$
where $_2F_1$ denotes the Hypergeometric function. To see this, note that we have that by the power-series definition for $_2F_1$, we have that:
$$\lim_{q\to 0}\frac{1}{q}(_2F_1(q,1;x+1;1)-1)=\sum_{i=1}^{\infty}\frac{(n-1)!}{\prod_{k=1}^{n}(x+k)}$$
On the other hand we have that:
$$\frac{\Gamma(x+1)\Gamma(x-q)}{\Gamma(x)\Gamma(x+1-q)}=\frac{x}{x-q}$$
and it is clear that:
$$\lim_{q\to 0}\frac{1}{q}(\frac{x}{x-q}-1)=\frac{1}{x}$$
Putting the sides together you get your identity.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\left.\sum_{n = 1}^{\infty}{\pars{n - 1}!\,x \over \prod_{k = 1}^{n}\pars{k + x}}
\,\right\vert_{\ x\ >\ 0}} =
x\sum_{n = 1}^{\infty}{\Gamma\pars{n} \over
\pars{1 + x}^{\large\overline{n}}}
\\[5mm] = &\
x\sum_{n = 1}^{\infty}{\Gamma\pars{n} \over
\Gamma\pars{1 + x + n}/\Gamma\pars{1 + x}} =
x\sum_{n = 1}^{\infty}{\Gamma\pars{n}\Gamma\pars{x + 1}\over
\Gamma\pars{n + x + 1}}
\\[5mm] = &\
x\sum_{n = 1}^{\infty}
\int_{0}^{1}t^{n - 1}\pars{1 - t}^{x}\dd t =
x\int_{0}^{1}{1 \over 1 - t}\pars{1 - t}^{x}\dd t
\\[5mm] = &\
x\int_{0}^{1}t^{x - 1}\,\dd t = \bbx{\large 1} \\ &
\end{align}
