Is it true that
$$(x+1)^{\bar{n}}= \sum_{k \ge 0} \sum _{i=0}^{n} {i \choose k}s_{n,i}\,\,x^k \,\,\,\,?$$ where $s_{n,i}$ is the Stirling number of the first kind and the $\bar{n}$ denote rising factorial. If yes, could anyone please show me?
I have been able to prove that $$(x+1)^{\bar{n}}= \sum_{k \ge 0} s_{n+1,k+1}\,\,x^k \,\,\,\,$$
My goal is to try to get that $$s_{n+1,k+1} = \sum _{i=0}^{n} {i \choose k}s_{n,i}$$ by equating (comparing) both identities
Are you familiar with the identity $$ x^{\overline n} = \sum_{i=0}^n s_{n,i}x^i?\tag{$*$} $$ The desired result follows pretty quickly substituting $x+1$ for $x$ in $(*)$: $$ (x+1)^{\overline n}=\sum_{i=0}^n s_{n,i}(x+1)^i = \sum_{i=0}^ns_{n,i}\sum_{k\ge 0}\binom{i}kx^k. $$ Identity $(*)$ can be proved by induction on $n$ using the relation $s_{n+1,k}=s_{n,k-1}+ns_{n,k}$, combined with $x^{\overline{n+1}}=x\cdot x^{\overline n}+nx^{\overline n}$. There is also a combinatorial proof.
For the identity
$$\sum_{p=k}^n {n\brack p} {p\choose k} = {n+1\brack k+1}$$
we obtain on the LHS using formal power series
$$n! [z^n] \sum_{p=k}^n {p\choose k} \frac{1}{p!} \left(\log\frac{1}{1-z}\right)^p$$
We have $\log\frac{1}{1-z} = z + \cdots$ so there is no contribution to the coefficient extractor when $p\gt n$ and we may continue with
$$n! [z^n] \sum_{p\ge k} {p\choose k} \frac{1}{p!} \left(\log\frac{1}{1-z}\right)^p \\ = n! [z^n] \sum_{p\ge 0} {p+k\choose k} \frac{1}{(p+k)!} \left(\log\frac{1}{1-z}\right)^{p+k} \\ = n! [z^n] \left(\log\frac{1}{1-z}\right)^{k} \sum_{p\ge 0} {p+k\choose k} \frac{1}{(p+k)!} \left(\log\frac{1}{1-z}\right)^{p} \\ = n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k} \sum_{p\ge 0} \frac{1}{p!} \left(\log\frac{1}{1-z}\right)^{p} \\ = n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k} \exp\log\frac{1}{1-z} \\ = n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k} \frac{1}{1-z}.$$
For the RHS we start from
$$\sum_{n\ge 0} {n\brack k+1} \frac{z^n}{n!} = \frac{1}{(k+1)!} \left(\log\frac{1}{1-z}\right)^{k+1}$$
so that by differentiation
$$\sum_{n\ge 0} {n+1\brack k+1} \frac{z^n}{n!} = \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k} \frac{1}{1-z}.$$
We see that the LHS and the RHS have the same EGF and this proves the claim. Here we have made repeated use of the labeled combinatorial class
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U}\times \textsc{CYC}(\mathcal{Z})).$$
This is the decomposition of permutations into sets of disjoint cycles.
The identity $$\sum _{p=k}^{n}{\left[{n \atop p}\right]{\binom {p}{k}}}=\left[{n+1 \atop k+1}\right]$$ can be proved by the techniques on the page Stirling numbers and exponential generating functions.
– Peter Taylor Sep 17 '18 at 12:57