Need some assistance on the expression $\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}$

Question

In this thread, Markus Scheuer attempted to prove that:

$$\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}=\frac{1}{1-x}\tag{1}$$

He wrote as followed:

$$\sum_{n=0}^\infty \frac{n!\color{red}{x^n}}{\color{green}{\prod_{j=1}^n(1+jx)}}=\sum_{n=0}^{\infty}\frac{n!}{\left(\color{blue}{1+\frac{1}{x}}\right)^{\overline{n}}}\tag{2}$$$$=\sum_{n=0}^{\infty}\frac{\color{purple}{1^{\overline{n}}1^{\overline{n}}}}{\left(1+\frac{1}{x}\right)}\,\color{ForestGreen}{\frac{1}{n!}}\tag{3}$$

where $q^{\overline{n}}=q(q+1)(q+2)...(q+n-1)$

I don't understand several points here:

1. First, on line $2$, why do $\color{red}{x^n}$ disspear? Did he multiply the denominator and numerator with $\dfrac{1}{x^n}$, if so, how can we multiply $\dfrac{1}{x^{n}}$ with the product $\prod_{j=1}^{n}(1+jx)$?, if so, isn't the product is going to be $\prod_{j=1}^{n}\left(\dfrac{1}{x^n}+\dfrac{j}{x^{1-n}}\right)$? I suspect that he factored the $x$ from the denominator, but then why $x^n$ disappear?

2. Second, how do you covert the product $\prod_{j=1}^{n}(1+jx)$ into $\left(1+\dfrac{1}{x}\right)^{\overline{n}}$? I try to write down to get a feel of it:

$\left(1+\dfrac{1}{x}\right)^{\overline{n}}=(1+\dfrac{1}{x})(2+\dfrac{1}{x})(3+\dfrac{1}{x})...$

The look similar to the product, but after factoring out $x$

3. Why can $\color{purple}{1^{\overline{n}}1^{\overline{n}}}$ replace $n!$ ? Also, why there is a new $\color{forestgreen}{\dfrac{1}{n!}}$

Answer 1

1. Yes, he did do that but no, notice that you have $n$ copies of $x$ i.e., $x^n,$ if you put the $x^n$ inside the product like that, then you will have $x^{n^2}$ so $n^2$ copies. Instead, put one each per term in the product, so $$\prod _{j=1}^n\left (\frac{1+jx}{x}\right )=\prod _{j=1}^n\left (\frac{1}{x}+j\right )$$
2. 1. solves this mystery.
3. Cause $x^{\overline{n}}=x(x+1)\cdots (x+n-1),$so replacing $x=1,$ you get $$1\cdot (1+1)\cdot (1+2)\cdots (1+n-1)=1\cdot 2\cdots (n-1)\cdot n=n!,$$ notice that there are two $1^{\overline{n}},$ so he multiplied and divide by $n!.$