Generating function of a polynomial Suppose I want to find a simple formula for the generating function of a general polynomial sequence $a_n=P(n)$.  Obviously it is enough to find the generating function of the sequence $a_n=n^k$ for any integer $k \geq 0$.  If we let $A_k(x) = \sum_{n=0}^{
\infty}n^kx^n$, it is not hard to see that: $$A_0(x)=\frac{1}{1-x}, A_k(x)=xA'_{k-1}(x)$$
My question is, it is possible from here to deduce an explicit formula for $A_k$?
 A: Define $$A_k(x) := \sum_{n=0}^{
\infty}n^kx^n \tag{1}$$
Then $$A_n(x) = \frac{P_n(x)}{(1-x)^{n+1}} \tag{2}$$
where the polynomials $\,P_n(x)\,$ are known as the
Eulerian polynomials with coefficients
given by the OEIS sequence A008292
"Triangle of Eulerian numbers T(n,k) (n>=1, 1<=k<=n) read
by rows." Many more details are contained in the OEIS entry.
For example, using the second formula in the entry gives an
explicit sum of $\,T(n,k)\,$ which yields the equation
$$ P_n(x) = \sum_{k=0}^n x^k\sum_{j=0}^k (-1)^j (k-j)^n {n+1\choose j}.
\tag{3} $$
You asked

My question is, it is possible from here to deduce an explicit formula for $A_k$?

and the answer is yes.
Suppose given $\,A_0(x)\,$ and define by recursion
$$ A_n(x) = x A'_{n-1}(x). \tag{4} $$
It can be shown that
$$ A_n(x) = \sum_{k=0}^n S(n,k)\, x^k A_0^{(k)}(x). \tag{5} $$
where $\,S(n,k)\,$ is the Stirling numbers of the second kind
(see OEIS sequence A008277).
In this particular case, $\,A_0(x) = \frac1{1-x}.\,$
If $\,B_0(x) := A_0(x),\;
B_n(x) := B\,'_{n-1} (x) = A_0^{(n)}(x),\,$ then it is
well-known that $\,B_n(x) = \frac{n!}{(1-x)^{n+1}}\,$ (see OEIS sequence A094587).
