I am trying to evaluate the following integral
$$\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_{n}, \hspace{0.5cm} 0<x<1$$
I tried to do this integral leveraging some polylogarithms but I am not making much progress.
Here is some of my work
$$\int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 = \int_{0}^{1-x_2} \frac{x_1}{1-x_1} \frac{1}{x_1} dx_1 = \int_{0}^{1-x_2} \frac{\text{Li}_0(x_1)}{x_1} dx_1 $$ $$ = \text{Li}_1(x_1)\Big|_{0}^{1-x_2} = \text{Li}_1(1-x_2),$$
where $\text{Li}_0$ is the $0^{th}$ polylogarithm. I use the relation $\text{Li}_m(x) = \int_{0}^{x} \frac{\text{Li}_{m-1}(x')}{x'} dx'$ and the fact $\text{Li}_m(0) = 0$.
Repeating the same kind of work
$$\int_{0}^{1-x_3} \frac{\text{Li}_1(1-x_2)}{1-x_2} dx_2 = - \int_{0}^{x_3} \frac{\text{Li}_1(u_2)}{u_2} du_2, \hspace{0.5cm} \text{where } u_2 = 1-x_2$$
$$\hspace{0.7cm} = \text{Li}_2(u_2)\Big|_{0}^{x_3} = \text{Li}_2(x_3)$$
I run into some problems at the third iteration
$$\int_{0}^{1-x_4} \frac{\text{Li}_2(x_3)}{1-x_3} dx_3 = \int_{0}^{1-x_4} \frac{\text{Li}_2(x_3)\text{Li}_0(x_3)}{x_3} dx_3 $$ $$ = \text{Li}_2(x_3)\text{Li}_1(x_3)\Big|_{0}^{1-x_4} - \int_{0}^{1-x_4}\frac{\text{Li}_1^2(x_3)}{x_3} dx_3$$
I've seen this last integral for the interval $[0,1]$ but I'm not sure how to extend it in such a way that I can get a nice recursion or sum representation.
Some relevant identities I stumbled upon while trying to solve this problem
$$\text{Li}_m(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^m}$$
$$\frac{\text{Li}_m(x)}{1-x} = \sum_{k=1}^{\infty} H_k^{(m)} x^k$$
Where $H_k^{(m)}$ is the generalized harmonic sum in powers $m$.
