TITLE: Partial fraction decomposition help
QUESTION [5 upvotes]: In a text that I am reading, they state that the following partial fraction ($r$ fixed) expansion is "readily computed":
$$f(z) = \frac{z^r}{(1-z)(1-2z)(1-3z)\cdots (1-rz)} = \frac{1}{r!} \sum_{j=0}^r \binom rj \frac{(-1)^{r-j}}{1-jz}$$
I know how to do partial fractions, or at least I thought I did. I tried to set it up with something like
$$\frac{A_1}{1-z} + \frac{A_2}{1-2z} + \cdots + \frac{A_r}{1-rz}$$
but it got messy. Also, the answer given above has a constant term $\frac{(-1)^{r}}{r!}$ when $j=0$, which shouldn't show up when doing this method normally.
Could someone point me in the right direction? Thanks!

I now see why the constant term is necessary; doing a "long division" would result in the $\frac{(-1)^r}{r!}$ term. The remainder would be something like
$$\frac{z^r - \frac{1}{r!}(1-z)(1-2z)\cdots(1-rz)}{(1-z)(1-2z)\cdots(1-rz)}$$
which technically could be solved by partial fractions...
Induction also looks promising, as suggested by Maesumi.

REPLY [3 votes]: The coefficient of $\frac{1}{1-kz}$ is
$$\left[\frac{z^r}{(1-z)(1-2z)(1-3z)\cdots\widetilde{(1-kz)}\cdots (1-rz)}\right]_{r=\frac{1}{z}}=\frac{\left(\frac{1}{k}\right)^r}{(1-\frac{1}{k})(1-2\frac{1}{k})(1-3\frac{1}{k})\cdots\widetilde{(1-k\frac{1}{k})}\cdots (1-r\frac{1}{k})}=\frac{1}{k(k-1)(k-2)\cdots\widetilde{0}(-1)(-2)\cdots(k-r)}=\frac{1}{r!}  \binom rj (-1)^{r-j}$$