Splitting $r^k (r+n)!$ as a sum of factorials I wanted to split the expression $r^k (r+n)!$ as a sum of factorials, where $k ,n \ \in \ \mathbb{Z} \ ;\ k>0$.
For example,

*

*$r(r+n)! = (r+n+1)! - (n+1)(r+n)!$


*$r^2(r+n)! = (r+n+2)! - (2n+3)(r+n+1)! + (n+1)^2 (r+n)!$
And in general,

If $$ r^k (r+n)! = \sum_{m=0}^{k} \lambda_{m} (r+n+m)! $$
where $ \lambda_{m} $ is independent of $r$, find a closed form expression for $\lambda_{m}$.


My Try :
1)  I tried to form a two variable recurrence for the expression and solve it such that we can express it in a summation form.
Let $r^k (r+n)! = A(k,n)$
Since $ r^k(r+n)! = r^{k-1}(\overline{r+n+1} - \overline{n+1})(r+n)! = r^{k-1}(r+n+1)! - r^{k-1} (n+1)(r+n)! $
$ \implies A(k,n) = A(k-1 , n+1) - (n+1)A(k-1,n) $
we can fix the initial conditions as $A(k,0) = r^k r!$ and $A(0,n) = (r+n)!$. I tried solving the above using generating functions, but eventually we'll have to find a Mac Lauren Series expansion for $\dfrac{1}{\Gamma \left( n+1 + \frac{2}{x} \right)}$
2) The expression can be rewritten as $$ r^k = \sum_{m=0}^{k} \lambda_{m} (r+n+m)_{m}  $$
where $(x)_{y}$ denotes the falling factorial. This resembles the identity $$ r^k = \sum_{m=0}^{k} {k\brace m} (r)_{m} $$
where $ \displaystyle {a\brace b} $ denotes the Stirling Numbers of the Second Kind. So maybe the the coefficients can be expressed in terms of Stirling Numbers.
3) I also tried plugging in values of $r$ and making a system of $k+1$ equations, but it became tedious to solve.

Update (26th November, 2016) : The user @Marko Riedel has discovered a remarkable closed form expression, namely
$$ \lambda_m = (-1)^{k+m}
\sum_{p=0}^{k-m} {k\choose p} {k+1-p\brace m+1} n^p $$
I have put a bounty since I feel that there might be alternate solutions to prove (or maybe even simplify) the above proposition. I'm also looking for any references/links to this problem.
Also, the great answers so far reveal that the identity is valid even when $n$ is not an integer.
Any help will be greatly appreciated.
 A: This answer is based upon an analysis of @MarkoRiedel's instructive answer. Although it  looks somewhat different it is essentially the same.
We consider OPs identity in the form
\begin{align*}
r^k=\sum_{m=0}^k\lambda_m \binom{n+r+m}{m}m!\tag{1}
\end{align*}

The essence
The binomial identity (1) in terms of corresponding generating functions is the relationship
  \begin{align*}
e^{-rz}=\frac{e^{(n+1)z}}{e^{(n+1+r)z}}\tag{2}
\end{align*}

We will  see that the generating function of $r^k$ is essentially the LHS of (2) whereas the generating function of the sum of the RHS   in    (1)  corresponds essentially to the coefficients of $z^n$ of the RHS of (2).

The claim
The following  is valid
  \begin{align*}
\lambda_m=(-1)^{m+k}\sum_{p=m}^k {p\brace m}\binom{k}{p}(n+1)^{k-p}\qquad\qquad    0\leq m\leq k
\end{align*}

In the following we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{r}{k}=[z^k](1+z)^r\qquad\text{and}\qquad  r^k=k![z^k]e^{rz}
\end{align*}

We obtain
  \begin{align*}
r^k&=(-1)^kk![z^k]e^{-rz}\\
&=(-1)^kk![z^k]\frac{e^{(n+1)z}}{e^{(n+1+r)z}}\tag{3}\\
&=(-1)^kk![z^k]\frac{e^{(n+1)z}}{\left(1-\left(e^z-1\right)\right)^{n+1+r}}\\
&=(-1)^kk![z^k]e^{(n+1)z}\sum_{m=0}^\infty\binom{-(n+1+r)}{m}(e^z-1)^m\tag{4}\\
&=(-1)^kk![z^k]e^{(n+1)z}\sum_{m=0}^k\binom{n+r+m}{m}(-1)^m(e^z-1)^m\tag{5}\\
&=(-1)^kk![z^k]e^{(n+1)z}\sum_{m=0}^k\binom{n+r+m}{m}(-1)^mm!\sum_{p=m}^\infty{p\brace m} \frac{z^p}{p!}\tag{6}\\
&=(-1)^kk!\sum_{m=0}^k\binom{n+r+m}{m}(-1)^mm!\sum_{p=m}^k{p\brace m} \frac{1}{p!}[z^{k-p}]e^{(n+1)z}\tag{7}\\
&=(-1)^kk!\sum_{m=0}^k\binom{n+r+m}{m}(-1)^mm!\sum_{p=m}^k{p\brace m} \frac{1}{p!}\cdot\frac{(n+1)^{k-p}}{(k-p)!}\tag{8}\\
&=\sum_{m=0}^k\binom{n+r+m}{m}m!\left((-1)^{m+k}\sum_{p=m}^k{p\brace m}\binom{k}{p}(n+1)^{k-p}\right)\tag{9}\\
\end{align*}
  and the claim follows.

Comment:


*

*In (3) we make an extension which looks simple but carries some power. In the following line we can rewrite the denominator to prepare for a binomial series expansion and it also contains the series for the Stirling numbers of the second kind.

*In (4) we apply the binomial series expansion.

*In (5) we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}
We also set the upper limit of the inner sum to $k$ by noting that the expansion of $(e^z-1)^m$ starts with $z^m$ and since we are looking for the coefficient of $z^k$ we can restrict the limit to $m=k$.

*In (6) we the use the following exponential series representation of the Stirling numbers of the second kind.
\begin{align*}
\sum_{p=m}^\infty{p\brace m} \frac{z^p}{p!}=\frac{(e^z-1)^m}{m!}
\end{align*}

*In (7) we do some rearrangements, apply the coefficient of operator rule
\begin{align*}
[z^{p-q}]A(z)=[z^{p}]z^qA(z)
\end{align*}
and restrict the upper limit of the inner sum to $k$ since the exponent $k-p$ has to be non-negative.

*In (8) we select the coefficient of $z^{k-p}$ of $e^{(n+1)z}$.

*In (9) we do some more rearrangements and collect terms to easier see $\lambda_m$.
A: The exposition herewith is not going to add anything  substantial to the valuable answers already 
provided by Markus and by Marko, but is intended just to show a couple of other ways to prove the assertion.
Clearly, for $r \in R$, $(r+n)!$ and $(r+n+k)!$ make sense only if represented through the $\Gamma$ function.
So, as already shown,  we actually are to prove that

$$
\begin{gathered}
  r^{\,q}  = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {\lambda _{\,q,\;j} \left( {r + n + j} \right)^{\,\underline {\,j\,} } }  = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {\lambda _{\,q,\;j} \left( {r + n + 1} \right)^{\,\overline {\,j\,} } }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {j!\lambda _{\,q,\;j} \left( \begin{gathered}
  r + n + j \\ 
  j \\ 
\end{gathered}  \right)}  = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {\left( { - 1} \right)^{\,j} j!\lambda _{\,q,\;j} \left( \begin{gathered}
 - r - n - 1 \\ 
  j \\ 
\end{gathered}  \right)} \quad \quad \left| \begin{gathered}
  \;r \in \mathbb{R}\,,\quad n \in \mathbb{Z} \hfill \\
  \;0 \leqslant q \in \mathbb{Z}\; \hfill \\
  \;\lambda _{\,q,\;j} \;\text{indep}\text{.}\,\text{from}\;r \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$

This is just an identity between polynomials, expressed in different basis,
so it is assured that a linear combination, with coefficients independent from $r$, exists and is unique. 


*

*0)  the straight one
$$
\begin{array}{l}
 r^{\,q}  = \sum\limits_{0\, \le \,j\, \le \,q} {\lambda _{\,q,\;j} \left( {r + n + 1} \right)^{\,\overline {\,j\,} } } \quad  \Rightarrow  \\ 
 \left( {s - \left( {n + 1} \right)} \right)^{\,q}  = \sum\limits_{0\, \le \,j\, \le \,q} {\lambda _{\,q,\;j} s^{\,\overline {\,j\,} } }  =  \\ 
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left\{ \begin{array}{c}
 q \\ 
 k \\ 
 \end{array} \right\}\left( {s - \left( {n + 1} \right)} \right)^{\,\overline {\,k\,} } }  =  \\ 
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left\{ \begin{array}{c}
 q \\ 
 k \\ 
 \end{array} \right\}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( \begin{array}{c}
 k \\ 
 j \\ 
 \end{array} \right)\left( { - \left( {n + 1} \right)} \right)^{\,\overline {\,k - j\,} } s^{\,\overline {\,j\,} } } }  =  \\ 
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left\{ \begin{array}{c}
 q \\ 
 k \\ 
 \end{array} \right\}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( { - 1} \right)^{\,k - j} \left( \begin{array}{c}
 k \\ 
 j \\ 
 \end{array} \right)\left( {n + 1} \right)^{\,\underline {\,k - j\,} } s^{\,\overline {\,j\,} } } } \quad  \Rightarrow  \\ 
 \end{array}
$$



$$
\lambda _{\,q,\;j}  = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\{ \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right\}\left( \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right)\left( {n + 1} \right)^{\,\underline {\,k - j\,} } }  = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\frac{{k!}}
{{j!}}\left\{ \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right\}\left( \begin{gathered}
  n + 1 \\ 
  k - j \\ 
\end{gathered}  \right)}    \tag{0}
$$



*

*1) : development of $r^q ==> (-r-n-1+n+1)^q$
The last of the four expressions formulated above suggests that we might try and develop $r^q$ as follows
$$
\begin{gathered}
  r^{\,q}  = \left( { - 1} \right)^{\,q} \left( { - r} \right)^{\,q}  = \left( { - 1} \right)^{\,q} \left( { - r - n - 1 + n + 1} \right)^{\,q}  = \left( { - 1} \right)^{\,q} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right)\left( {n + 1} \right)^{\,q - k} \left( { - r - n - 1} \right)^{\,k} }  =  \hfill \\
   = \left( { - 1} \right)^{\,q} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right)\left( {n + 1} \right)^{\,q - k} \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,k} \right)} {j!\left\{ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right\}\left( \begin{gathered}
 - r - n - 1 \\ 
  j \\ 
\end{gathered}  \right)} }  \hfill \\ 
\end{gathered} 
$$
which compared with the starting identity gives:


$$
\lambda _{\,q,\;j}  = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right)\left\{ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right\}\left( {n + 1} \right)^{\,q - k} }  = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left( \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right)\left( {n + 1} \right)^{\,q - k} \left( { - 1} \right)^{\,k - j} \left\{ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right\}}  \tag{1}
$$

which is the answer anticipated by Markus Scheuer.


*

*2) Eulerian Numbers and Worpitzky identity
The target identity is quite similar to the Worpitzky's Identity, except
for having the binomial diagonally shifted. So it is not difficult to convert into the standard Worpitzky form
$$
r^{\,q}  = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered}
  q \hfill \\
  k \hfill \\ 
\end{gathered} 
 \right\rangle } \left( \begin{gathered}
  r + k \\ 
  q \\ 
\end{gathered}  \right) = \; \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered}
  q \hfill \\
  k \hfill \\ 
\end{gathered}  \right\rangle } \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered}
   - n + k - 1 - j \\ 
  q - j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r + n + j \\ 
  j \\ 
\end{gathered}  \right)} 
$$
where the Eulerian Numbers can be expressed in different ways, among which
$$
\begin{gathered}
  \left\langle \begin{gathered}
  q \\ 
  m \\ 
\end{gathered}  \right\rangle  = \sum\limits_{0\, \leqslant \,k\, \leqslant \,m} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  q + 1 \\ 
  k \\ 
\end{gathered}  \right)\left( {m + 1 - k} \right)^{\,q} }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,q - m} \right)\,} {\left( { - 1} \right)^{\,q - m + k} \left( \begin{gathered}
  q + 1 \\ 
  m + 1 + k \\ 
\end{gathered}  \right)\,k^{\,q} }  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( { - 1} \right)^{\,q - k - m} \left\{ \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right\}\left( \begin{gathered}
  q - k \\ 
  m \\ 
\end{gathered}  \right)\,\;k!}  \hfill \\ 
\end{gathered} 
$$
So this leads to express $\lambda _{\,q,\;j} $ as


$$
\lambda _{\,q,\;j}  = \frac{1}
{{j!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered}
  q \hfill \\
  k \hfill \\ 
\end{gathered}  \right\rangle } \left( \begin{gathered}
   - n + k - 1 - j \\ 
  q - j \\ 
\end{gathered}  \right) = \frac{1}
{{j!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered}
  q \hfill \\
  k \hfill \\ 
\end{gathered}  \right\rangle } \left( { - 1} \right)^{\,q - j} \left( \begin{gathered}
  q + n - k \\ 
  q - j \\ 
\end{gathered}  \right)  \tag{2}
$$

In conclusion let's note that the identities above could be conveniently put into
matricial notation, whereby one can more easily get the properties  of $\lambda$.
For instance identity (1) can be written as:

$$
\mathbf{\Lambda }(n) = \mathbf{B}^{\, - \,\mathbf{(n + 1)}} \;\mathbf{St}_{\,\mathbf{1}} ^{\, - \,\mathbf{1}}  = \;\left( {\mathbf{St}_{\,\mathbf{1}} \,\mathbf{B}^{\,\,\mathbf{(n + 1)}} } \right)^{\, - \,\mathbf{1}} \quad  \Rightarrow \quad \Lambda _{\,q,\,m} (n) = \left. {\frac{1}
{{q!}}\nabla _{\,x} ^m \,x^{\,q} } \right|_{\,x\, = \, - \,(n + 1)}  \tag{3}
$$  

where all matrices are LT (indexed from $0$), and
$\mathbf{B}$ is the Pascal matrix
$\mathbf{St}_{\,\mathbf{1}}$ is the matrix of the unsigned Stirling N. of 1st kind
$\nabla_{\,x}$ is the backward Delta
so that the last identity indicates the connection with the Newton backward development of $r^q$,
that comes of no surprise.
Given the wide range of properties of the Pascal matrix, the formula above allows to transfer them to
$\mathbf{\Lambda }(n) \mathbf{St}_{\,\mathbf{1}}$ and $ \mathbf{St}_{\,\mathbf{1}} \mathbf{\Lambda }(n)$.  
Finally, it shall be remarked that the exposition made extends quite evenly to  $r$ and $n$ complex.


*

*----   Addendum -----*  


Might be interesting to signalize that, for $n$ non-negative integer, $\lambda$ can be expressed in terms
of the so called r-Stirling Numbers ( re. e.g. to this paper by A. Z. Broder ),
denoted as $\left\{ \begin{gathered}  n \\   m \\ \end{gathered}  \right\}_r $.
Accordingly, formula (1) can be rewritten as 

$$
\begin{gathered}
  \lambda _{\,q,\;j} (n) = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered}
  q \\ 
  k \\ 
\end{gathered}  \right)\left\{ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right\}\left( {n + 1} \right)^{\,q - k} } \quad \left| {\;\;0 \leqslant n \in \;\;\mathbb{Z}\,} \right.\quad  =  \hfill \\
   = \left( { - 1} \right)^{\,q - j} \left\{ \begin{gathered}
  q + n + 1 \\ 
  j + n + 1 \\ 
\end{gathered}  \right\}_{n + 1}  = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n + 1} \right)} {\left( { - 1} \right)^{\,k} \left[ \begin{gathered}
  n + 1 \\ 
  n + 1 - k \\ 
\end{gathered}  \right]\left\{ \begin{gathered}
  q + n + 1 - k \\ 
  n + 1 + j \\ 
\end{gathered}  \right\}}  \hfill \\ 
\end{gathered}   \tag{a}
$$  

thereby getting an additional expression in terms of standard Stirling Numbers.
($\left[ \begin{gathered}  n \\   m \\ \end{gathered}  \right]$ being the unsigned Stirling N. of 1st kind).
A: Start with
$$
\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
\begin{align}
r^k
&=(-1)^k((n+1)-(r+n+1))^k\tag{1}\\[9pt]
&=(-1)^k\sum_{j=0}^k\binom{k}{j}(-r-n-1)^j(n+1)^{k-j}\tag{2}\\
&=\sum_{j=0}^k\sum_{m=0}^j(-1)^k\binom{k}{j}(n+1)^{k-j}\binom{-r-n-1}{m}\stirtwo{j}{m}m!\tag{3}\\
&=\sum_{m=0}^k\color{#C00000}{\sum_{j=m}^k(-1)^{k-m}\binom{k}{j}(n+1)^{k-j}\stirtwo{j}{m}}\binom{r+n+m}{m}m!\tag{4}
\end{align}
$$
Explanation:
$(1)$: rewrite $r=-((n+1)-(r+n+1))\vphantom{\sum\limits_{m=0}^j}$
$(2)$: apply the Binomial Theorem
$(3)$: $x^j=\sum\limits_{m=0}^j\binom{x}{m}\stirtwo{j}{m}m!$
$(4)$: change order of summation and use $\binom{-r-n-1}{m}=(-1)^m\binom{r+n+m}{m}$
Thus, if we set
$$
\lambda_m=(-1)^{k-m}\sum_{j=m}^k\binom{k}{j}(n+1)^{k-j}\stirtwo{j}{m}\tag{5}
$$
then $(4)$ becomes
$$
r^k=\sum_{m=0}^k\lambda_m\binom{r+n+m}{m}m!\tag{6}
$$
which is equivalent to your statement when $r+n\in\mathbb{Z}$.
A: We can rewrite the condition $\enspace\displaystyle\sum\limits_{m=0}^k\lambda_{k,m}(r+n+m)! =r^k (r+n)!\enspace$ as $$\sum\limits_{m=0}^k \lambda_{k,m} (x+m)^{\underline{m}}=(x-n)^k $$ by substituting $\enspace r\enspace $ with $\enspace x-n$ . 
Using the method of differences of Discrete Mathematics with $\enspace\Delta f(x):=f(x+1)-f(x)\enspace$ for 
any polynomial $ f(x)$, $\enspace\displaystyle \Delta^{n+1}:=\Delta(\Delta ^n)\enspace$ and therefore 
first $\enspace\displaystyle \Delta^n f(x)= \sum\limits_{j=0}^n (-1)^{n-j}\binom{n}{j}f(x+j)\enspace$ and second $\enspace\displaystyle \Delta^l (x+m)^{\underline{m}} =m^{\underline{l} }(x+m)^{ \underline{m-l} }$ 
for $\enspace m\geq l\enspace $ and $\enspace \Delta^l (x+m)^{\underline{m}}=0\enspace $ for $\enspace m<l\enspace $ it follows with $\enspace f(x):=(x-n)^k\enspace$ : 
$$\sum\limits_{m=l}^k \lambda_{k,m} m^{\underline{l} }(x+m)^\underline{m-l}= \Delta^l \sum\limits_{m=0}^k \lambda_{k,m} (x+m)^\underline{m}=$$ $$=\Delta^l (x-n)^k =\sum\limits_{j=0}^l (-1)^{l-j}\binom{l}{j}(x-n+j)^k$$
With $\enspace x:=-(l+1) \enspace $ and therefore $\enspace (x+m)^\underline{m-l}=0\enspace $ for $\enspace m>l\enspace $ and deviding the equation by $\enspace l! \enspace $ we get $$\lambda_{k,l}=\frac{(-1)^k}{l!}\sum\limits_{j=0}^l (-1)^j\binom{l}{j}(n+1+j)^k \enspace .$$
Hint: 
$\lambda_{k,l}:=\lambda_{k,l}(n)\enspace $ with $\enspace n\in\mathbb{C}\enspace $ is one of several useful generalizations of the Stirling numbers of the second kind $\enspace \displaystyle{k\brace l}:=(-1)^{k+l}\lambda_{k,l}(-1) \,$ . 
