From the generating function for the function $x/\log(1+x)$, denoting $(A_n)_{n\geq 0}$ the corresponding sequence of coefficients, by integration of the function $1/\log(t+1)$ over $ \left( 1,x-1 \right) $, after the change of variable, $u=t+1$ ($u$ is relabeled after as $t$ one more time) one has
$$\text{Li}(x)=\int_1^{x-1}\frac{1}{t}A_0dt+\int_1^{x-1}\frac{1}{t} \left(\sum_{n=1}^\infty\frac{A_n}{n!}t^n\right)dt= $$
where I take the definition $\text{Li}(x):=\int_2^x \frac{dt}{\log t}$. Then I can use (if there are no mistakes) that $A_0:=f^{(0)}(0)\equiv f(0)=\frac{0}{\log 1}=\frac{0}{0}$, and L'Hôpital's rule assures me that this $A_0$ is $\lim_{x\to 0}\frac{x}{\log(1+x)}=\lim_{x\to 0} 1+x=1$. By a change of the index of summation,
$$=1\cdot\log(x-1)+\int_1^{x-1}\sum_{n=0}^{\infty}\frac{A_{n+1}}{(n+1)!}t^n\cdot(-1)^n\cdot(-1)^n dt.$$
Then using the second identity deduced in PROBLEMA 53 La Gaceta de la RSME Volumen 10, Número 1, in Spanish (page 155), we can obatin the logarithmic integral written in terms of Bernoulli numbers, binomial numbers, and Stirling numbers of the first kind (those defined by brackets)
$$\text{Li}(x)=\log(x-1)-\sum_{n=0}^\infty(-1)^{n+1}\sum_{k=0}^n\frac{(-1)^k\left((x-1)^{n+1}-1\right)}{(n+1)!}\begin{bmatrix} n\\ k \end{bmatrix}\frac{B_{k+1}}{k+1}.$$
Also I know how to expand and simplify $(x-1)^{n+1}-1$ using Newton's binomial theorem, but I believe that today I was sufficiently artificious, and I believe that it isn't possible to get the sum of the sums in the previous identity.
Question. Was there any mistake in my computations? If yes, please say where and provide your own final formula. If my computations were right, take this as a proof verification and tell me if there are some inaccuracies or additional justifications that you want to notify to me. Thanks in advance.