Proof of Identity to Zero of the Sum of a Product of Binomial Coefficients & Pochhammer Numbers It's well-know that the sum across an entire row of binomial coefficients (of degree, say, $n$) with alternating signs attached is 0; and it can easily be proven by demonstrating that it is the binomial expansion of $(1-1)^n$. What is less obvious is that if you take the same series & multiply the terms in it by consecutive Pochhammer numbers of degree $n-1$, but from anywhere along that series, the sum is still zero! To put it formally $$\sum_{k=0}^n \frac{(-1)^k(m+k)_{n-1}}{k!(n-k)!}=0 ,$$ $∀m\geq n-1$, it not mattering that the $n!$ is omitted in the denominator, as only the relative size of the terms matters, the sum being putatively 0. That this is so follows from a tracing of the combinatorial consequence on the terms in the series expansion about 1 of the logarithm of the property of the logarithm, that the logarithm of a product = the sum of the logarithms; so if we have a proof through some other route (which we do) that the property & the series imply each other, then we effectively already have a proof of the theorem adduced as the grounds of this post ... but an exceedingly roundabout proof! There must be a more elementary proof than that ... but I can't figure it out. I have a very vague recollection of seeing one when reading elementary stuff on binomial coefficients a long time ago; but I'm not absolutely certain I did - it might have been a proof of some other theorem.
It did occur to me that one route through which it might be proven that of recasting the $n$th row (the very top one with $1$ only in it being the zeroth) of the binomial coefficient triangle as being the series $$1+\sum_{k=0}^{n-1}\prod_{l=0}^k\frac{l-k}{l+1} ,$$ whence the one queried in this post being recast as  $$1+\sum_{k=0}^{n-1}\prod_{l=0}^k\frac{(l-k)(l+m+n)}{(l+1)(l+m+1)} ∀m∊ℕ_0$$ the absolute value mattering not atall, as it is only the relative value of the terms that matters, the sum being putatively zero.
I don't know that this is the best way to the proof, but it did occur to me it might be viable. I wonder whether anyone would show a proof or a clue to one, whether by this means or otherwise.
There might possibly be a clue in the fact that the lastmentioned series could just as well be $$1+\sum_{k=0}^{n-1}\prod_{l=0}^k\frac{(l-k)(l+m+1)}{(l+1)(l+m+n)} ∀m∊ℕ_0$$ as because the binomial coefficients are symmetrical from front to back, it matters not in which direction the series of Pochhammer numbers is traversed.
 A: The forward Finite Difference of a function $f(x)$ is defined as
$$
\Delta \,f(x) = f(x + 1) - f(x)
$$
and its iteration as
$$
\Delta ^{\,n} \,f(x) = \Delta \left( {\,\Delta ^{\,n - 1} \,f(x)} \right)
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{n - k} \left( \matrix{ n \cr  k \cr}  \right)\;f(x + k)} 
$$
It is a known fact that if $f(x)=p_q(x)$ is a polynomial of degree $q$ then, similarly to derivatives, 
$$
\Delta ^{\,n} \,p_{\,q} (x) = 0\quad \left| {\;q < n} \right.
$$
So the case you propose is just
$$
\eqalign{
  & \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^k \;} {{\left( {x + k} \right)_{\,n - 1} } \over {k!\left( {n - k} \right)!}}
 = \left( { - 1} \right)^n n!\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{n - k} \left( \matrix{
  n \cr 
  k \cr}  \right)\;(x + k)_{n - 1} } _\,  =   \cr 
  &  = \left( { - 1} \right)^n n!\Delta ^{\,n} \,x^{\,\overline {\,n - 1\,} }
  = \left( { - 1} \right)^n n!\Delta ^{\,n} \prod\limits_{0\, \le \,j\, \le \,n - 2} {\left( {x + j} \right)} \, = 0 \cr} 
$$
with x integer, real or complex.
A: Starting from
$$\sum_{k=0}^n (-1)^k \frac{1}{k!(n-k)!} (m+k)^\underline{n-1}$$
we get
$$\sum_{k=0}^n (-1)^k \frac{(n-1)!}{k!(n-k)!}
{m+k\choose n-1}
= \frac{1}{n} \sum_{k=0}^n (-1)^k {n\choose k}
{m+k\choose n-1}
\\ = \frac{1}{n} \sum_{k=0}^n (-1)^k {n\choose k}
[z^{n-1}] (1+z)^{m+k}
\\ = [z^{n-1}] (1+z)^{m}
\frac{1}{n} \sum_{k=0}^n (-1)^k {n\choose k} (1+z)^k
\\ = [z^{n-1}] (1+z)^{m}
\frac{1}{n} (1-(1+z))^n
\\ = \frac{(-1)^n}{n} [z^{n-1}] z^n (1+z)^{m}
= 0.$$
Here we have  used the fact that  $z^n (1+z)^{m} = z^n  + \cdots$ note
however that we are extracting the coefficient on $[z^{n-1}].$ Careful
inspection reveals  that this  will go  through for  $n\ge 1$  and all
integer $m$  including negative. This  is because $(1+z)^m$  is entire
when $m\ge  0$ and  the only  pole of  $(1+z)^m$ when  $m\lt 0$  is at
$z=-1.$
Observe  that  this  is  the combinatorial  interpretation  of  the
Pochhammer symbol as the falling factorial  as opposed to the one from
special     functions.      These     two    are     documented     at
MathWorld.   The
proof for the alternate notation is left to the reader.
