Calculating "factorial sequence" of a rational function

Dividing by $k!$ we write $$ f(k)=c_0+c_1\cdot \frac1k+c_2\cdot \frac1{k(k-1)}+\ldots+c_{N-1}\frac1{k(k-1)\ldots(k-N+2)}+O(k^{-N}). $$ Denoting $k=1/x$ this reads as $$ f(1/x)=c_0+c_1x+c_2\frac{x^2}{1-x}+c_3\frac{x^3}{(1-x)(1-2x)}+\ldots+c_{N-1}\frac{x^{N-1}}{(1-x)(1-2x)\ldots(1-(N-2)x)}+O(x^N), \quad x\to 0. $$ If $f(k)=p(k)/q(k)$ and $\deg p\leqslant \deg q$, we may expand $f(1/x)$ as a power series in $x$ and find $c_0,c_1,\ldots$ consequently equalizing the coefficients of $1,x,x^2,\ldots$ in both sides.

A comment on the issue of determining the coefficients of the expansion in function series in Fedor Petrov's answer. Recall the generating function of the Stirling numbers of the second kind $S(n,r)$ : $$\phi_r(x):=\frac{x^r}{(1-x)\dots(1-rx)}=\sum_{n\ge0}S(n,r)x^n,$$ and the "Stirling inversion" (i.e. the fact that the infinite triangular matrices of the Stirling numbers of the first and second kind are inverses of each other), that can be expressed as $$x^n=\sum_{n\ge0} s(r,n)\phi_r(x).$$

As a consequence, any formal $\phi_r$ series can be expanded in a formal power series and conversely,
$$\sum_{r\ge0} c_r\phi_r=\sum_{n\ge0}b_nx^n,$$ the correspondence being: $$ b_n=\sum_{r=0}^n S(n,r)c_r$$ $$ c_r=\sum_{n=0}^r s(r,n)b_n.$$

Warning. These notations differ slightly from your expansion, that is by a shift of indices and by a factor $x$, which can be easily adjusted. In fact, if you do not have strong objections to change notation, it could be worth to adapt it to that of Stirling numbers.

Let's assume that $p$ is monic of degree $d$ and has distinct roots $r_1,\dots,r_d$. Consider the partial fraction decomposition: $$f(x) = \frac{1}{p(x)} = \sum_{i=1}^d \frac{a_i}{x-r_i},$$ where $a_i:=\frac{1}{p'(r_i)}$.

Then $$\frac{1}{x}f(\frac{1}{x}) = \sum_{i=1}^d \frac{a_i}{1-r_ix}.$$ From answers of Fedor and Pietro, and the generating function for Stirling numbers of first kind, it follows that $$\sum_{n\geq 0} c_n \frac{z^n}{n!} = \sum_{i=1}^d a_i (1+z)^{r_i},$$ and thus $$c_n = n! \sum_{i=1}^d a_i \binom{r_i}n = \sum_{i=1}^d a_i (r_i)_n.$$

It is straightforward to further generalize this to the case of repeated roots and/or $f$ being a ratio of two polynomials.