Combinatorial proof of $x^{(n)} = \sum_{k = 1}^n L(n,k)(x)_k$ Question: Using a combinatorial proof show the following identity.
$$x^{(n)} = \sum_{k = 1}^n L(n,k)(x)_k,$$
where $x^{(n)}$ denotes the rising factorial and $(x)_k$ denotes the falling factorial.
I am also intersted in why it suffices to show such formulas only for $x \in \mathbb{N}$ and then expect them to hold for real or even complex $x$.
Context:
We know that, $L(n,k)$ are Lah numbers satisfiying this recurrence relation: $$L(n,k) = L(n-1, k-1) + (n - 1 + k)L(n - 1,k),$$
and this explicit formula $L(n,k) = \frac{n!}{k!}\binom{n-1}{n-k}$. I have seen a few pages where Lah numbers were defined as the connecting coefficients between the rising and falling factorials, which is what I am trying to show, but I have yet to find a proof of my desired statement.
 A: Noticing that $\binom{a}{b}=\frac{(a)_b}{b!}, $ you can write your expression as
$$\binom{x+n-1}{n}=\sum _{k=1}^n\binom{n-1}{n-k}\binom{x}{k}$$ which is Vandermonde, so the standard combinatorial proof of Vandermonde suffices. Notice that the explicit formula comes from shuffling the elements of $[n]$ in a line, using stars and bars to see that $\binom{n-1}{n-k}$ is the number of ways to partition the $n$ into positive $k$ parts $a_1+a_2+\cdots +a_k=n$ where order matters, and so you partition your line using first $a_1$ numbers, then $a_2$ until you get the $n$ numbers, and taking out the order of the $k$ parts. For example: Given $a_1=4,a_2=2,a_3=1,a_4=4$ then
$$\underbrace{1\,9\,10\,2}_{a_1}\,\underbrace{5\,6}_{a_2}\,\underbrace{4}_{a_3}\,\underbrace{7\,3\,8}_{a_4}\text{ gives the ordered partition.}$$
It suffices to show it just for $x$ an integer because this are polynomials on $x$ and if you have two polynomials $P_1(x)=P_2(x)$ for $x\in \mathbb{N}$ then $$P_1(x)-P_2(x)=0$$ implies that the polynomial on the LHS has infinite roots, that is just possible if the polynomial on the left is strictly $0.$
