A combinatoric identity involving Eulerian and Stirling numbers Let
$$
A_{n}(x)=\sum_{k=0}^{n}\left\langle\begin{array}{l}
n \\
k
\end{array}\right\rangle x^{n-k}
$$
Eulerian Polynomials
and $$
\left\langle\begin{array}{l}
n \\
k
\end{array}\right\rangle
$$ are Eulerian numbers.
I want to ask whats the proof of this interesting equation?
$$
\sum_{k=1}^{n}\left\langle\begin{array}{l}
n \\
k
\end{array}\right\rangle x^{n-k}=(1-x)^{n} \sum_{k=0}^{n}\left\{\begin{array}{l}
n \\
k
\end{array}\right\} k !\left(\frac{x}{1-x}\right)^{k}
$$
I have a reference https://www.mat.uniroma2.it/~tauraso/AMM/AMM11007.pdf
In this paper(American Mathematical Monthly Problem)  there are a similar approach. It solves 2 seperate equation with induction.You can look at it.
When i solved this equation in same way ,i couldnt reach a recurence relation about eulerian and stirling numbers. Maybe i go wrong in somewhere.
So can you help me ?
 A: Here is another proof for the curious. We seek to verify that
(with $n\ge 1$, $n=0$ holds by inspection)
$$\sum_{k=0}^n \left\langle n \atop k \right\rangle x^{n-k}
= (1-x)^n \sum_{k=0}^n {n\brace k} k!
\left(\frac{x}{1-x}\right)^k.$$
We get using standard EGFs for the RHS
$$n! [z^n] (1-x)^n
\sum_{k=0}^n \frac{(\exp(z)-1)^k}{k!} k!
\left(\frac{x}{1-x}\right)^k
\\ = n! [z^n] (1-x)^n
\sum_{k=0}^n (\exp(z)-1)^k
\left(\frac{x}{1-x}\right)^k.$$
Now because $\exp(z)-1 = z+\cdots$ we have $(\exp(z)-1)^k = z^k +
\cdots$ so when $k\gt n$ there is no contribution to the coefficient
extractor and we get
$$n! [z^n] (1-x)^n
\sum_{k\ge 0} (\exp(z)-1)^k
\left(\frac{x}{1-x}\right)^k
\\ = n! [z^n] (1-x)^n
\frac{1}{1-(\exp(z)-1)x/(1-x)}
\\ = n! [z^n] (1-x)^n
\frac{1-x}{1-x-(\exp(z)-1)x}
\\ = n! [z^n] (1-x)^n
\frac{1-x}{1-x \exp(z)}
\\ = n! [z^n] \frac{1-x}{1-x\exp(z(1-x))}.$$
On the other hand we have for the LHS by the mixed GF of the Eulerian
numbers
$$n! [z^n] \sum_{k=0}^n x^{n-k}
[w^k] \frac{w-1}{w-\exp((w-1)z)}$$
Now we have $\left\langle n \atop k \right\rangle = 0$ when $k\ge n$ so
this is
$$n! [z^n] x^n \sum_{k\ge 0} x^{-k}
[w^k] \frac{w-1}{w-\exp((w-1)z)}
\\ = n! [z^n] x^n \frac{1/x-1}{1/x-\exp((1/x-1)z)}
\\ = n! [z^n] x^n \frac{1-x}{1-x\exp((1/x-1)z)}
\\ = n! [z^n] \frac{1-x}{1-x\exp((1/x-1)zx)}
\\ = n! [z^n] \frac{1-x}{1-x\exp((1-x)z)}.$$
The LHS is the same as the RHS and we have the claim.
Addendum. We have
$$n! [z^n] [w^k] \frac{w-1}{w-\exp((w-1)z)}
\\ = n! [z^n] [w^{k+1}] \frac{w-1}{1-\exp((w-1)z)/w}
\\ = n! [z^n] [w^{k+1}] (w-1)
\sum_{q\ge 0} \frac{1}{w^q} \exp(q(w-1)z)
\\ = [w^{k+1}]
\sum_{q\ge 0} \frac{1}{w^q} q^n (w-1)^{n+1}
= \sum_{q\ge 0} [w^{k+1+q}] q^n (w-1)^{n+1}
\\ = (-1)^{n-k} \sum_{q=1}^{n-k} (-1)^q q^n {n+1\choose k+1+q}.$$
This justifies that $\left\langle n \atop k \right\rangle = 0$ when $k\ge
n$ and hence the two coefficient extractors combined return  zero in that
case as  claimed.
