I would like to show the formula
$$\forall n \in \mathbb{N},\quad\sum_{p=0}^n \binom{n}{p} \frac{(-1)^p}{p!} \prod_{k=0}^{p-1} \left(n+\frac{1}{2} + k\right) = \binom{-1/2}{n} \, , $$
which is true according to Mathematica. I tried to use the identity
$$\binom{-1/2}{n} = \frac{1}{n!} \sum_{k=0}^n s_{n,k}\left(-\frac{1}{2}\right)^k \, ,$$
where $s_{n,k}$ denotes the Stirling numbers of first kind, but without any success. We end up with the following identity to prove
$$ \frac{1}{n!} \sum_{p=k}^n \binom{n}{p}\frac{s_{p,k}}{p!} \left(n+\frac{1}{2}\right)^k = s_{n,k} \, , $$
which is not a piece of cake for me neither. Among all the formulas I could find in Concrete Mathematics book from Ronald Graham, Donald Knuth, and Oren Patashnik, none comes in a handy. I confess I don't have a large culture of this field.
I will be thankfull for any hint.
Best Answer
More than a hint: Recall that the falling factorial is defined as $x^\underline{k}=\frac{x!}{(x-k)!}=x(x-1)\cdots (x-k+1),$ and the rising factorial is defined as $x^\overline{k}=\frac{(x+k-1)!}{(x-1)!}=x(x+1)\cdots (x+k-1).$
This is the binomial theorem $$(x+y)^n=\sum _{k=0}^n\binom{n}{k}x^ky^{n-k}.$$ The beautiful part is that this keeps happening for $$(x+y)^{\underline{n}}=\sum _{k=0}^n\binom{n}{k}x^{\underline{k}}y^{\underline{n-k}},$$ notice that if you take $x=n,y=-1/2-n,$ then $$(-1/2)^{\underline{n}}=\sum _{p=0}^n\binom{n}{p}n^{\underline{n-p}}(-1/2-n)^{\underline{p}}.$$ Then you have to check that $(-x)^{\underline{k}}=(-1)^k(x)^{\overline{k}}.$