The generating function for the Eulerian polynomials is
$$\frac{t-1}{t-e^{(t-1)x}} = \sum_{n=0}^\infty A_n(x) \frac{t^n}{n!}$$ where $A_n(x)$ is the $n^{th}$ Euler polynomial and $$A_n(x) = \sum_{k=0}^n e_{nk}t^k $$.
Question: how can I prove that the below equality is also true?
$$(*)\ \ \ \ \ \frac{1-t}{1-te^{(1-t)x}} = \sum_{n=0}^\infty A_n(x) \frac{t^n}{n!}$$
The Eulerian polynomials have been defined in more than one way in the past.
Some define $A_n(x)=\sum_{w\in\mathfrak{S}_n}x^{\mathrm{des}(w)}$ (see, for example, Shareshian, Wachs).
This gives the generating function as $$\frac{t-1}{t-e^{(t-1)x}} = \sum_{n=0}^\infty A_n(x) \frac{t^n}{n!}.$$
Others (for example, Foata) define the Eulerian polynomial to be generated by the statistic $(1+\mathrm{des}(w))$ for all $n\geq 1$, i.e. $A_n(x)=\sum_{w\in\mathfrak{S}_n}x^{1+\mathrm{des}(w)}$.
This will have the generating function that you have described later. $$\frac{1-t}{1-te^{(1-t)x}} = \sum_{n=0}^\infty A_n(x) \frac{t^n}{n!}$$ As noted by Joriki and Alexander Burstein, these are not the same.
See this wonderful this article on the evolution of the Eulerian polynomial from Euler's time to the present by Foata.