Generating polynom for stirling numbers of second kind - Where's my mistake? Let $S(a,b)$ denote the Stirling numbers of the second kind, i.e. the number of partitions of a set with $a$ elements into $b$  non-empty subsets.
Let further denote 
$$
F_k\left(x\right)=\sum_{n\geq0}{S\left(k,n\right)x^n}
$$
the generating function for the Stirling numbers, where we set $k$ as a fixed parameter.
We immediately get that 
$$
F_k\left(x\right)=\sum_{n\geq0}{S\left(k,n\right)x^n}=\sum_{n\geq0}^{k}{S\left(k,n\right)x^n}
$$
, as we can't partition a set into more subsets than it has elements.
We start with $\sum_{n\geq0}{S\left(k,n\right)x^n}$ and substitute into it the recurrence $S(a,b) = S(a-1,b-1)+k\cdot S(a-1,b)$:
$$
\begin{align*}
&\sum_{n\geq0}{S\left(k,n\right)x^n}=\left(\sum_{n\geq0}{S\left(k-1,n-1\right)x^n}\right)+k\cdot\sum_{n\geq0}{S\left(k-1,n\right)x^n}\\&\Leftrightarrow\\&F_k\left(x\right)=\left(\sum_{n\geq0}{S\left(k-1,n-1\right)x^n}\right)+k\cdot F_{k-1}\left(x\right)\\&\Leftrightarrow\\&F_k\left(x\right)=\left(\sum_{n\geq1}{S\left(k-1,n-1\right)x^n}\right)+k\cdot F_{k-1}\left(x\right)\ +\ S\left(k-1,-1\right)x^0\\&\Leftrightarrow\\&F_k\left(x\right)=x\left(\sum_{n\geq0}{S\left(k-1,n\right)x^n}\right)+k\cdot F_{k-1}\left(x\right)\ \\&\Leftrightarrow\\&F_k\left(x\right)=xF_{k-1}\left(x\right)+k\cdot F_{k-1}\left(x\right)\\&\Leftrightarrow\\&F_k\left(x\right)=\left(x+k\right)\cdot F_{k-1}\left(x\right)
\end{align*}
$$
Using the initial value $F_0(x) = 1$ we obtain the generating function:
$$
F_k(x) =\sum_{n\geq0}{S\left(k,n\right)x^n}= \prod_{i=1}^{k}\left(x+i\right)
$$
So, if I e.g. set $k=5$ I should obtain:
$$
\prod_{i=1}^{5}\left(x+i\right)  = S(5,0) + S(5,1)x + ... + S(5,5)x^5
$$
However, this is wrong for $n=0$ already, as
$$
[x^0]\prod_{i=1}^{5}\left(x+i\right) = 5!
$$
and not $0$.
Where's my mistake?

After correcting (both the starting index as pointed out by John, as well as the mistake in the application of the recursion) it becomes:
$$
\begin{align*}
&\left(\sum_{n\geq1}{S\left(k,n\right)x^n}\right)+S\left(k,0\right)=\left(\sum_{n\geq1}{\left(S\left(k-1,n-1\right)+nS\left(k-1,n\right)\right)x^n}\right)+S\left(k,0\right)\\&\Leftrightarrow\\&\left(\sum_{n\geq0}{S\left(k,n\right)x^n}\right)=\left(\sum_{n\geq1}{S\left(k-1,n-1\right)x^n}\right)+\left(\sum_{n\geq1}{\left(nS\left(k-1,n\right)\right)x^n}\right)+S\left(k,0\right)\\&\Leftrightarrow\\&\left(\sum_{n\geq0}{S\left(k,n\right)x^n}\right)=x\left(\sum_{n\geq0}{S\left(k-1,n\right)x^n}\right)+\left(\sum_{n\geq1}{\left(nS\left(k-1,n\right)\right)x^n}\right)+S\left(k,0\right)\\&\Leftrightarrow\\&F_k\left(x\right)=xF_{k-1}\left(x\right)+\left(\sum_{n\geq0}{\left(nS\left(k-1,n\right)\right)x^n}\right)+S\left(k,0\right)\\&\Leftrightarrow\\&F_k\left(x\right)=xF_{k-1}\left(x\right)+x\frac{d}{dx}\left(\sum_{n\geq0}{\left(S\left(k-1,n\right)\right)x^n}\right)+S\left(k,0\right)\\&\Leftrightarrow\\
&F_k\left(x\right)=xF_{k-1}\left(x\right)+x\frac{d}{dx}F_{k-1}\left(x\right)+S\left(k,0\right)
\end{align*}
$$
This now leads to a combination of a recursion with a ODE... which is frankly too much for me.
 A: A basic mistake is you're using the recurrence relation outside of the range it's defined for. As the Recurrence relation section of Wikipedia's "Stirling numbers of the second kind" article states

Stirling numbers of the second kind obey the recurrence relation
     $\left\{{n+1 \atop k}\right\}=k\left\{{n \atop k}\right\}+\left\{{n \atop k-1}\right\}$
  for $k \gt 0$ ...

Using your notation and equation, this becomes

$$S(a,b) = S(a-1,b-1)+k\cdot S(a-1,b) \tag{1}\label{eq1}$$

which is only applicable for $b \gt 0$. However, you're using it starting with $b = 0$ (i.e., $n = 0$) in your next steps.
I believe a roughly comparable situation is the recurrence relation for factorials. You have that $0! = 1$ and, for $n \gt 0$,
$$n! = n(n-1)! \tag{2}\label{eq2}$$
If you try to go to before that, e.g., $0! = 0(-1)!$, you get that $(-1)! = \frac{1}{0}$, which can obviously cause issues if you're using it to try to prove something else. However, I'm not saying the same particular issue occurs here, but just that you can't necessarily trust the behavior of recurrence relations if you use them outside of their stated bounds.
