The Lah numbers count ordered-block set partitions and expand the rising factorial in the falling basis

Fix $n\ge 1$ and $k\in \{1,\dots ,n\}$. Take a permutation of $[n]$, written as a word of length $n$, and choose $k-1$ of the $n-1$ gaps between consecutive letters. Cutting the word at those gaps produces an ordered list of $k$ nonempty ordered blocks. This gives $n!\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)$ ordered lists of ordered blocks.

Forgetting the left-to-right order of the $k$ blocks divides by $k!$, because every unordered family of $k$ internally ordered blocks has exactly $k!$ linear orders of its blocks. Hence $L(n,k)$ counts partitions of $[n]$ into $k$ nonempty linearly ordered blocks.

For $m\in \mathbb{N}$, consider $m$ distinguishable boxes arranged from left to right. Building an ordered list inside each box by inserting the elements $1,2,\dots ,n$ one after another gives $m(m+1)\cdots (m+n-1)=m^{\bar{n}}$ possibilities. Grouping the outcomes by the number $k$ of nonempty boxes, one first chooses the underlying partition of $[n]$ into $k$ internally ordered blocks, counted by $L(n,k)$ from step 2.1, and then chooses the $k$ occupied boxes in order, which gives $m^{\underset{‾}{k}}$ possibilities. Summing over $k$ yields $m^{\bar{n}}={\sum }_{k=0}^{n}L(n,k)m^{\underset{‾}{k}}$.

Steps 2.1 and 3.1 prove the counting interpretation and the falling-factorial expansion on natural arguments, together with the defining value $L(0,0)=1$.