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?

Using @Andrew Chin's comments, we face for the second series
$$S_\infty=\sum_{n=0}^\infty\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}$$
Let $$a_n=\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}=\frac{\Gamma (n+1)}{\left(1+\frac{1}{x}\right)_n}$$ where appear Pochhammer symbols. So
$$S_p=\sum_{n=0}^p a_n=\frac{\frac{\Gamma (p+2)\, \Gamma \left(\frac{1}{x}\right)}{\Gamma
   \left(p+1+\frac{1}{x}\right)}-1}{x-1}$$ and, assuming $0< x <1$
$$\lim_{p\to \infty } \, \frac{\Gamma (p+2) \Gamma \left(\frac{1}{x}\right)}{\Gamma
   \left(p+1+\frac{1}{x}\right)}=0$$ since, using expansion for large $p$
$$\frac{\Gamma (p+2) \Gamma \left(\frac{1}{x}\right)}{\Gamma
   \left(p+1+\frac{1}{x}\right)}=\Gamma \left(\frac{1}{x}\right) p^{1-\frac{1}{x}}\left(1+ \frac{(x-1) (2 x+1)}{2 x^2}\frac 1p+O\left(\frac{1}{p^2}\right)\right)$$
