Stirling Numbers and partial fraction decomposition I wanted to ask how can I break into partial fraction this expression:
$$ \frac {1}{(1-x)(1-2x)...(1-kx)}$$
I tried saying for $x=\frac{1}{i}$ we get some expression for $A_i$ but got stuck.
Thank you very much!
 A: $$f(x)=\frac{1}{\prod_{k=1}^{n}(1-kx)}=\sum_{k=1}^{n}\frac{A_k}{x-\frac{1}{k}}, $$
$$ A_k = \text{Res}\left(f(x),x=\frac{1}{k}\right) = \frac{-1}{k\prod_{\substack{1\leq j\leq n\\ j\neq k}}\left(1-\frac{j}{k}\right)}$$
$$A_k = \frac{-k^{n-1}}{k\prod_{\substack{1\leq j\leq n\\ j\neq k}}(k-j)}=\color{red}{\frac{k^{n-1}(-1)^{n-k+1}}{k!(n-k)!}}.$$
A: As the poles are all simple we can use partial fractions by residues.
With 
$$f(z) = \prod_{k=1}^n \frac{1}{1-kz}$$
we have
$$f(z) = \sum_{k=1}^n \frac{1}{z-1/k} \mathrm{Res}_{z=1/k} f(z).$$
To extract these residues we write
$$f(z) = \frac{1}{n!} \prod_{k=1}^n \frac{1}{1/k-z}
= \frac{(-1)^n}{n!} \prod_{k=1}^n \frac{1}{z-1/k}$$
We then get for the residue
$$\mathrm{Res}_{z=1/k} f(z) =
\frac{(-1)^n}{n!} \prod_{j=1, j\ne k}^n \frac{1}{1/k-1/j}
\\ = \frac{(-1)^n}{n!} \prod_{j=1, j\ne k}^n \frac{kj}{j-k}
= \frac{(-1)^n}{n!} k^{n-1} \prod_{j=1, j\ne k}^n \frac{j}{j-k}
\\ = \frac{(-1)^n}{n!} k^{n-1} 
\frac{n!}{k} \frac{(-1)^{k-1}}{(k-1)! (n-k)!}
= (-1)^{n+k-1} k^{n-1} \frac{1}{k! (n-k)!}.$$
