Set $F_k(x) := \sum_{n\geq k} S(n,k)x^n$. Prove that $$F_1(x) = \frac{x}{1-x}, \space \space \space F_2(x) = \frac{x^2}{(1-x)(1-2x)} $$
Furthermore, show that the function $F_k(x)$ satisfy the recurrence relation $$F_k(x) = \frac{x}{1-kx}F_{k-1}(x)$$ and solve this recurrence.
Any ideas how to approach this problem?
I prefer the notation $n\brace k$ for your $S(n,k)$. The first step is to observe that ${n\brace 1}=1$ for $n\ge 1$: there is only one partition of $[n]=\{1,\ldots,n\}$ into just one part. Thus,
$$F_1(x)=\sum_{n\ge 1}{n\brace 1}x^n=\sum_{n\ge 1}x^n=x\sum_{n\ge 1}x^{n-1}=x\sum_{n\ge 0}x^n=\frac{x}{1-x}\;.$$
To get $F_2(x)$, we need to figure out what $n\brace 2$ is. Suppose that we partition $[n]$ into two non-empty parts. One of the parts must contain the number $n$; call that part $A$ and the other part $B$. What can the rest of $A$ look like? It can’t be all of $[n-1]$, since then $B$ would be empty, but it can be any other subset of $[n-1]$. Since $[n-1]$ has $2^{n-1}$ subsets, there are $2^{n-1}-1$ possible choices for the rest of $A$ and hence for a $2$-part partition of $[n]$. Thus, ${n\brace 2}=2^{n-1}-1$ for $n\ge 2$, and
$$\begin{align*} F_2(x)&=\sum_{n\ge 2}{n\brace 2}x^n\\ &=\sum_{n\ge 2}\left(2^{n-1}-1\right)x^n\\ &=\sum_{n\ge 2}2^{n-1}x^n-\sum_{n\ge 2}x^n\\ &=2x^2\sum_{n\ge 0}(2x)^n-x^2\sum_{n\ge 0}x^n\\ &=\frac{2x^2}{1-2x}-\frac{x^2}{1-x}\\ &=\frac{2x^2(1-x)-x^2(1-2x)}{(1-x)(1-2x)}\\ &=\frac{x^2}{(1-x)(1-2x)}\;. \end{align*}$$
At this point we want the standard recurrence relation satisfied by the Stirling numbers of the second kind,
$${n\brace k}=k{{n-1}\brace k}+{{n-1}\brace{k-1}}\;,$$
substituting it into the definition of $F_k(x)$:
$$\begin{align*} F_k(x)&=\sum_{n\ge k}{n\brace k}x^n\\ &=k\sum_{n\ge k}{{n-1}\brace k}x^n+\sum_{n\ge k}{{n-1}\brace{k-1}}x^n\\ &=kx\sum_{n\ge k-1}{n\brace k}x^n+x\sum_{n\ge k-1}{n\brace{k-1}}x^n\\ &=kxF_k(x)+xF_{k-1}(x)\;. \end{align*}$$
Now just solve for $F_k(x)$ in terms of $F_{k-1}(x)$.
