Show that the number of ways of splitting a set with $n$ elements into $k$ non-empty subsets is equal to $x^n$'s coefficient in a series

Question

The series evaluates to (could not put it in the title): $$\frac{x^{k}}{(1-x)(1-2x)...(1-kx)}$$ I tried to come up with something. A solution would probably require a use of the geometric series formula as well as Cauchy's product. The task also demands one to deduce the following formula for the Stirling numbers of the second kind:

$$S(n, 3) = \frac{3^{n-1} - 2^{n} + 1}{2}$$ which does show up as a coefficient after multiplying: $$({\sum_{n = 0}^{\infty}x^{n}})({\sum_{n = 0}^{\infty}(2x)^{n}})({\sum_{n = 0}^{\infty}(3x)^{n}}).$$ I guess I should be able to follow from there, yet I am stuck.

Edit: Thank you for the link as well as the proof. Is it possible to proceed from my suggestion about using the geometric series formula?

Answer 1

Let $S(n,k)$ be the number of partitions requested. Let $R(n,k)$ denote the coefficient of $x^n$ in the given series $F_k(x)$. Clearly $S(0,0)=1= R(0,0)$ and for $n,k>0$ we have $S(n,0)=0=R(n,0)$ and $S(0,k)=0=R(0,k)$. One way to generate a partition of $\{1,...,n+1\}$ into $k$ nonempty sets is to make $\{n+1\}$ a singleton and partition $\{1,...,n\}$ into $k-1$ nonempty sets. The only other way is to start with partition of $\{1,...,n\}$ into $k$ nonempty sets, and adjoin $n+1$ to one of these $k$ sets. This gives the recurrence formula $$S(n+1,k)=S(n,k-1)+kS(n,k) \,. \quad (*)$$ On the other hand $$F_k(x)=\sum_{j=1}^{\infty} k^{j-1}x^j F_{k-1}(x)$$ implies that $$R(n,k)=\sum_{j=1}^{n+1-k} k^{j-1}R(n-j,k-1) \,. \; \quad (**)$$ Writing $i=j+1$ we have $$R(n,k-1)+kR(n,k)=R(n,k-1)+\sum_{i=2}^{n+2-k} k^{i-1}R(n+1-i,k-1) =R(n+1,k)\,,$$ where the rightmost equality follows from (**) with $n$ replaced by $n+1$. Comparing this to (*) we conclude by induction on $n$ that $R(n,k)=S(n,k)$ for all $n,k$.