How to derive this series 
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.

I know the first series and it is easy to obtain it. But the second series is strange. It is not a power series, not a Taylor series. How does one obtain this series?

Proof of the Formula
Below, we show inductively
$$
\frac1{1-x}=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag1
$$
where the empty sum is $0$ and the empty product is $1$.
Gautschi's Inequality says that
$$
\begin{align}
\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}
&=\frac{\Gamma(n+1)\,\Gamma\!\left(1+\frac1x\right)}{\Gamma\!\left(n+1+\frac1x\right)}\\
&\sim\frac{\Gamma\!\left(1+\frac1x\right)}{(n+1)^{1/x}}\tag2
\end{align}
$$
Thus, for $0\lt x\lt1$, the series
$$
\sum_{k=0}^\infty\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}\tag3
$$
converges and the remainder term
$$
\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag4
$$
vanishes as $n\to\infty$. Therefore, for $0\lt x\lt1$,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^\infty\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}=\frac1{1-x}}\tag5
$$

Inductive Proof of $\bf{(1)}$
Trivially, we have that $(1)$ is true for $n=0$.
Suppose that we have $(1)$ for some $n$. Then
$$
\begin{align}
\frac1{1-x}
&=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=0}^n(1+jx)}\frac{1+nx}{1-x}\\
&=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=0}^n(1+jx)}\left(\color{#C00}{1}+\color{#090}{\frac{(n+1)x}{1-x}}\right)\\
&=\sum_{k=0}^{\color{#C00}{n}}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\color{#090}{\frac{(n+1)!\,x^{n+1}}{\prod_{j=1}^{n+1}(1+jx)}\frac{1+(n+1)x}{1-x}}\tag6
\end{align}
$$
Thus, $(1)$ holds for $n+1$.
