I would like to prove that: $$f_m(x) = \dfrac{x^m}{(1-x)(1-2x)...(1-mx)}$$
Where $$f_m(x) = \sum_{n=0}^{\infty} S(n,m)x^n$$ and $S(n,m)$ is stirling number of 2nd kind
Multiplying the recurrence relation $S(n,m) = S(n-1,m-1) + m*S(n-1,m) $ by $x^n$ and simplifying (skipping some steps) I get:
$$f_m(x) = \dfrac{x}{1-mx}f_{m-1}(x)$$ The next step is what I am not sure of: $$f_m(x)= (\dfrac{x}{1-mx})^m$$
$$ f_m(x) = \dfrac{x^m}{(1-x)(1-2x)...(1-mx)}$$
Is this correct? Or am I completely wrong?
You’re right to have doubts about that step: $f_{m-1}(x)$ isn’t $\left(\frac{x}{1-mx}\right)^{m-1}$, so from
$$f_m(x)=\frac{x}{1-mx}f_{m-1}(x)\tag{1}$$
you cannot conclude that
$$f_m(x)=\left(\frac{x}{1-mx}\right)^m\;.$$
(Note that this also doesn’t give you what you want. However, if you assume as an induction hypothesis that
$$f_{m-1}(x)=\prod_{k=1}^{m-1}\frac{x}{1-kx}\;,$$
then $(1)$ would let you conclude that
$$f_m(x)=\prod_{k=1}^m\frac{x}{1-kx}=\frac{x^m}{(1-x)(1-2x)\ldots(1-mx)}\;.$$
Alternately, you can use the identity that $$S(n, k) = \frac{1}{k!}\sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^n.$$ Then is you apply the Norlund-Rice integral formula referenced here, you can obtain this generating function by residue calculus.
