1. Recall the Stirling numbers of the 2nd kind are $S(n,k) := \# \text{ set partitions of } \{1,2,..., n\} \text{ into }$ \newline $k$ (non-empty) blocks. We saw $F_k(x) := \sum_{n\ge 0} S(n,k)x^n$ satisfies $F_k(x) = \frac{x^k}{(1-x)(1-2x)...(1-kx)}$ \newline for any $k \ge 1$. Find the partial fraction decomposition of $F_k(x)$, i.e., find the numbers $a_j \in \mathbb{Q}$ \newline for which $F_k(x) = a_0 + \frac{a_1}{(1-x)} + \frac{a_2}{(1-2x)} + ... + \frac{a_k}{(1-kx)}$. Conclude that $S(n,k) = \sum_{j=0}^k a_j j^n$. \newline Hint: Clear denominators, and then plug in $x = \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{k}$ and finally $x = 0$.
