How can we evaluate this $\prod_{k=1}^n(1+kx)$ $\displaystyle\prod_{k=1}^n(1+kx)=\underbrace{\displaystyle\sum_{k=0}^n a_k x^k}_{\text{I assumed this,it don't have to be like this}}$
I'm  investigating what this means, how we can analyse this and get     generalized formula.
In fact ,I thought $n-$degree equaliton's formulas.
For instance ,assume this $\displaystyle\prod_{k=1}^n(1+kx)=a_0+a_1x+....+a_{n-1}x^{n-1}+a_nx^n$
And I think we know $\displaystyle\sum \left(\dfrac{-1}{k_i}\right)=-\dfrac{a_{n-1}}{a_n}$
It's like , when  ($ ax^2+bx+c=0 $)  , $x_1+x_2=\dfrac{-b}{a}$
And I kept doing this , but this was gonna last to eternity...

Notice that
$$\prod_{k=1}^n(1+kx)=\prod_{k=0}^n(1+kx)=1\times(1+x)\times(1+2x)\times\dots\times(1+nx)\\=(1+nx)\overbrace{!!\dots!!}^x\text{ or }(1+nx)!^{(x)}$$
where the long string of exclamation marks is the multifactorial, the extension of the double factorial.  The Wikipedia of the factorial also has a small section on this.
As the Wikipedia states, this can be rewritten as follows:
$$\prod_{k=1}^n(1+kx)=x^n\frac{\Gamma(n+1+\frac1x)}{\Gamma(1+\frac1x)}$$
where $\Gamma$ is the Gamma function.  This form allows extension to any $n,x\in\mathbb C$.
