Combinatorial (and Algebraic Proof) of an Identity Involving Lah Numbers
A partial combinatorial proof may go as follows:
The set $S=\{1,2,\ldots , n+1\}$ can be split into $k+1$ non-empty ordered sets in $L_{n+1,k+1}$ ways.
Alternatively, element $k+1$ is in an ordered set on it's own in $L_{n,k}$ ways or appears between elements in the $k+1$ ordered sets in $(n+k+1)L_{n,k+1}$ ways.
Hence the recurrence
$$L_{n+1,k+1}=L_{n,k}+(n+k+1)L_{n,k+1}\, .$$
Iterating:
$$\begin{align}L_{n+1,k+1}&=L_{n,k}+(n+k+1)(L_{n-1,k}+(n+k)L_{n-1,k+1})\\[1ex] &=L_{n,k}+(n+k+1)L_{n-1,k}+(n+k+1)(n+k)L_{n-1,k+1}\\[1ex] &=L_{n,k}+(n+k+1)L_{n-1,k}+(n+k+1)(n+k)(L_{n-2,k}+(n+k+1)L_{n-2,k+1})\\[1ex] &=L_{n,k}+(n+k+1)^\underline{1}L_{n-1,k}+(n+k+1)^\underline{2}L_{n-2,k}+(n+k+1)^\underline{3}L_{n-2,k+1}\\[1ex] &\hspace1.4ex\vdots\\[1ex] &=\sum_{i=0}^{n-k}(n+k+1)^\underline{i}L_{n-i,k}\, .\end{align}$$
$L_{n+1,k+1}$ is the number of partitions of $\{1,2,\dots,n,n+1\}$ into $k+1$ non-empty ordered subsets. Let the rank of such a partition be the largest number $i$ for which $i+1$ is not in the same subset as any of $1,2,\dots,i$. Then
$L_{i,k}(n+k+1)_{n-i}$ is the number of partitions with rank $i$.
Why? A partition of rank $i$ can be specified by first choosing a partition of $\{1,2,\dots,i+1\}$ into ordered subsets, then adding in the elements $\{i+2,i+3,\dots,n+1\}$. The elements $\{1,2,\dots,i+1\}$ must span all $k+1$ subsets, since by definition of rank, every subsequent element is in a subset with a previous element. Since $i+1$ is in a different subset than $\{1,2,\dots,i\}$, there are $L_{i,k}$ ways to place the numbers $\{1,2,\dots,i+1\}$. Then, the element $i+2$ can be added to one of these subsets in $k+i+2$ ways (either at the start of one of the $k+1$ subsets, or immediately after one of the $i+1$ elements). Next, element $i+3$ can be added in $k+i+3$ ways, then $i+4$ can be added in $k+i+4$ ways, and so on.
Putting this all together, the number of ways to choose the partition of rank $i$ is $$ L_{i,k}\cdot (k+i+2)\cdot(k+i+3)\cdot\ldots\cdot(n+k+1)=L_{i,k}(n+k+1)_{n-i}. $$
Side note: the combinatorial proof in your post leads to a proof of the similar identity $$ L_{n+1,k+1}=\sum_i L_{i,k}\binom{n}i(n+1-i)!=\sum_i L_{i,k}(n)_{n-i}\cdot(n+1-i). $$