Rising Factorial and Stirling number of the 1st kind 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
 A: 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.
A: 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.
