We shall show an upper bound on the Stirling number of the second kind, a byproduct of a homework exercise of Probabilistic Combinatorics offered by Prof. Tom Bohman.

**Definition** A Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of \(n\) objects into \(k\) non-empty subsets and is denoted by \(S(n,k)\).

**Proposition** For all \(n, k\), we have $$S(n,k) \leq \frac{k^n}{k!}\left(1-(1-1/m)^k\right)^m.$$

**Proof** Consider a random bipartite graph with partite sets \(U:=[n], V:=[k]\). For each vertex \(u\in U\), it (independently) connects to exactly one of the vertices in \(V\) uniformly at random. Suppose \(X\) is the set of non-isolated vertices in \(V\). It is easy to see that $$\operatorname{Pr}\left(X=V\right) = \frac{\text{number of surjections from }U\text{ to }V}{k^n} = \frac{k!S(n,k)}{k^n}.$$

On the other hand, we claim that for any \(\emptyset \neq A \subset [k]\) and \(i \in [k]\setminus A\), $$\operatorname{Pr}\left(i\in X \mid A\subset X\right) \leq \operatorname{Pr}\left(i\in X\right).$$ Note that the claim is equivalent to $$\operatorname{Pr}\left(A\subset X \mid i\notin X\right) \geq \operatorname{Pr}\left(A\subset X\right).$$ Consider the same random bipartite graph with \(V\) replaced by \(V’:=[k]\setminus \{i\}\) and let \(X’\) be the set of non-isolated vertices in \(V’\). The claim is justified since $$\operatorname{Pr}\left(A\subset X\mid i\notin X\right) = \operatorname{Pr}\left(A\subset X’\right) \geq \operatorname{Pr}\left(A\subset X\right).$$

Set \(A:=[i-1]\) in above for \(i = 2, \ldots, k\). Using the multiplication rule with telescoping the conditional probability, we obtain $$\begin{eqnarray}\operatorname{Pr}\left(X=V\right) &=& \operatorname{Pr}\left(1\in X\right)\operatorname{Pr}\left(2\in X \mid [1]\subset X\right)\ldots \operatorname{Pr}\left(k\in X\mid [k-1]\subset X\right)\\ & \leq & \operatorname{Pr}\left(1\in X\right)\operatorname{Pr}\left(2\in X\right)\ldots\operatorname{Pr}\left(k\in X\right) \\ & = & \left(1-(1-1/m)^k\right)^m.\end{eqnarray}$$ QED.

4 (this post is made with love)