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Prof. Putinar,
Guillera & Sondow gave
for x\geq 0, to which I add
for \mbox{Re} <u>\geq 0</u>.
-Ben Orin
Guillera & Sondow gave
e^{x}=\prod_{n=1}^{\infty}\left(\prod_{k=1}^{n} (1+kx)^{(-1)^{k+1}\left(\begin{array}{c}n\\k\end{array}\right)} \right) ^{\frac{1}{n}}
for x\geq 0, to which I add
\boxed{\frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n} (k+u)^{(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) (k+u)} \right) ^{\frac{1}{n+1}}}
for \mbox{Re} <u>\geq 0</u>.
-Ben Orin