For a positive integer $n$ choose $m$ with $1 \le m \le n$ and let $X$ be a set with $|X|=n$. choose $x_j \in X$ for $j=1,...,m$.
Define $m$ directed graphs $X_j$ in the following way:
Initially we have for each $j$: $$ X_j =\{x_j\} $$ if (as sets) $$ \cup_j X_j = X $$ then we are done. otherwise choose $x \in X-\cup_j X_j$ and choose $k$ with $1 \le k \le m$. take $y \in X_k$ and add $y$ to the nodeset of $X_k$. then add a single arc directed from $y$ to some other element of $X_k$.
at the termination of this procedure we have partitioned $X$ into disjoint sets $X_j$ and have given each $X_j$ the additional structure of a directed tree with root $x_j$.
Question how many ways are there of doing this?
Motivation: this came out of thinking about the structure of endomorphisms of a finite set. for a partition into trees along the lines indicated, every permutation $\sigma \in S_m$ of the set $\{x_j\}$ allows us to define an endomorphism of X as follows: $ \forall x_j $ add the arc joining $x_j$ to $\sigma(x_j)$. every node of the resulting graph now has out-degree $1$.
My hypothesis is that every endomorphism may be represented as a unique graph as described. thus if $T_{n,m}$ is the number of way of partitioning a set of $n$ elements into $m$ directed trees, then we have: $$ \sum_{m=1}^n T_{n,m} m! = n^n $$