Showing that $x^n=\sum_{k=1}^{n}{n\brace k}(x)(x-1)\ldots (x-k+1)$ holds for all numbers, not just positive integers I just finished proving that this statement holds for all positive integers $r$ (through a combinatorial argument) $$r^n=\sum_{k=1}^{n}{n\brace k}(r)(r-1)\ldots (r-k+1)$$ (where the curly braces indicate the Stirling set numbers) However, I'm being asked to show that this holds for all numbers $x$. That is $$x^n=\sum_{k=1}^{n}{n\brace k}(x)(x-1)\ldots (x-k+1)$$ The hint we received was that we should note that the second equation has at most $n$ solutions. However, I'm not sure how that will help me here. Any help would be greatly appreciated.
 A: I’ll expand on the hint. Write the second equation in the form
$$f(x)=x^n-\sum_{k=1}^n{n\brace k}\prod_{i=0}^{k-1}(x-i)\;;$$
$f(x)$ is a polynomial, and you’ve shown that $f(r)=0$ for each $r\in\Bbb Z^+$. What polynomial has infinitely many zeroes?
A: Suppose we seek to verify that
$$x^n = \sum_{k=1}^n {n\brace k} x^{\underline{k}}$$
for all $x$ including $x$ complex.
We start with the basic
$$x^{\underline{k}} = \sum_{q=1}^k 
\left[k\atop q\right] (-1)^{k+q} x^q.$$
This  follows from  the  defining recurrence  of  the signed  Stirling
numbers of the first kind and certainly holds for any $x.$
Substitute this into the target sum to get
$$x^n = \sum_{k=1}^n {n\brace k} 
\sum_{q=1}^k 
\left[k\atop q\right] (-1)^{k+q} x^q
\\ = \sum_{q=1}^n x^q
\sum_{k=q}^n {n\brace k} \left[k\atop q\right] (-1)^{k+q}.$$
We thus have to show that
$$\sum_{k=q}^n {n\brace k} \left[k\atop q\right] (-1)^{k+q}
= \begin{cases} 1 \quad\text{if}\quad q=n 
\\ 0 \quad\text{if}\quad 1\le q\lt n.\end{cases}.$$
The first of these follows by inspection. For the second
recall the species for set partitions which is
$$\mathfrak{P}(\mathcal{U} \mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1)).$$
and the species of cycle decompositions
$$\mathfrak{P}(\mathcal{U} \mathfrak{C}_{\ge 1}(\mathcal{W}))$$
which gives the generating function
$$H(w, u) = \exp\left(u\left(\log\frac{1}{1-w}\right)\right).$$
Substitute these into the sum to get
$$n! [z^n] \sum_{k=q}^n \frac{(\exp(z)-1)^k}{k!}
k! [w^k] (-1)^{k+q} 
\frac{1}{q!} \left(\log\frac{1}{1-w}\right)^q.$$
Now the exponential term makes  a zero contribution to the coefficient
extractor  at  the  front when  $k\gt  n$  so  we  may extend  $k$  to
infinity.  The logarithmic  term  does not  contribute  to the  second
coefficient extractor when $k\lt q$ so  we may start $k$ at zero. This
yields
$$n! (-1)^q [z^n] \sum_{k\ge 0} (\exp(z)-1)^k
(-1)^{k} [w^k]
\frac{1}{q!} \left(\log\frac{1}{1-w}\right)^q.$$
Now what we have here in $w$ is an annihilated coefficient extractor
(ACE) that simplifies to
$$n! (-1)^q [z^n]
\frac{1}{q!} \left(\log\frac{1}{1-(1-\exp(z))}\right)^q
= n! (-1)^q [z^n] \frac{1}{q!} (-z)^q.$$
This is zero unless $q=n$ when it is one as claimed and we are done.

Remark. If there are any  issues with the last simplification here
it can be  verified using formal power series only  and the proof is
at this MSE link.

Remark, II.  There are several  more examples of the  technique of
annihilated    coefficient   extractors    at    this   MSE    link
I  and at  this MSE
link  II  and  also
here             at             this             MSE             link
III.
