I'd like to prove the following identity:
$$\sum_{n=0}^{\infty}S\left(n,k\right)\,x^{n}=\frac{x^{k}}{(1-x)(1-2x)\cdots(1-% kx)}$$
(The identity referred from : http://dlmf.nist.gov/26.8#E11)
Rather than using induction, I would like to find another way to prove this identity.
I just superficially understands that the generating function has a intrinsic relation to recurrence relation, and what I know about the recurrence upon Stiring the second kind is:
$$S(n,k) = S(n-1,k-1) + k \cdot S(n-1,k)$$
How can I make up the object identity from this recurrence relation?
We know that if $(n,k) \neq (0,0)$ then we have $S(n, k)= S(n-1, k-1)+kS(n-1,k)$ and we define $S(0,0)=1.$
Let $B_k(x)= \displaystyle \sum_{n\geq 0}S(n, k)x^n$ so multiplying the recurrence above by $x^n$ and summing over $n\geq 0$ we get
$B_{k}(x)= xB_{k-1}(x)+kxB_k(x)$
$\Rightarrow B_k(x)= \dfrac{xB_{k-1}(x)}{1-kx}=\dfrac{x^2B_{k-2}(x)}{(1-kx)(1-(k-1)x)}=\ldots= \dfrac{x^k}{\displaystyle\prod_{r=1}^k(1-rx)}$ thereby establishing the identity.