I am interested in proving that 


$$\sum\_{k=0}^n\frac{k}{k!}\sum\_{l=0}^{n-k}\frac{(-1)^l}{l!}=1 
$$
and 
$$\sum\_{k=0}^n\frac{k^2}{k!}\sum\_{l=0}^{n-k}\frac{(-1)^l}{l!}=2. 
$$
I verified it numerically for many values of $n$ and would like now to get to a proof. I rewrote it to make incomplete Gamma functions appear, but it is (apparently) not leading me further. 


Has somebody ever encountered such an identity ? Is there an easy proof I am just completely missing ? I apologize if the question is trivial, I may lack some combinatorics tools. 


Note: It was edited and should now be correct



Sums of binomial coefficients weighted by incomplete gamma
Let $\mathbb{N}$ be the set $\left\{ 0,1,2,\ldots\right\} $.


If $n\in\mathbb{N}$ and $k\in\mathbb{N}$, then we shall use the notation $\genfrac{\{}{\}}{0pt}{0}{n}{k}$ for the number of set partitions of the set $\left\{ 1,2,\ldots,n\right\}
$ into $k$ nonempty subsets. This is a [Stirling number of the 2nd
kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind).
We will mainly use the following formula:
\begin{equation}
\genfrac{\{}{\}}{0pt}{0}{n}{k}=\dfrac{1}{k!}\sum\_{i=0}^{k}\left( -1\right) ^{i}\dbinom{k}{i}\left(
k-i\right) ^{n}.
\label{darij1.eq.stir=}
\tag{1}
\end{equation}
For proofs of \eqref{darij1.eq.stir=}, see pretty much any text on enumerative
combinatorics. For example, \eqref{darij1.eq.stir=} is the first equality sign
of the formula (19) in [David Galvin, *Basic discrete mathematics*, version 13
December 2017](http://www.cip.ifi.lmu.de/~grinberg/t/17f/60610lectures2017-Galvin.pdf).


The definition of $\genfrac{\{}{\}}{0pt}{0}{n}{k}$ yields two facts:


* The number $\genfrac{\{}{\}}{0pt}{0}{n}{k}$ is a nonnegative integer for any $n\in\mathbb{N}$ and $k\in\mathbb{N}$.
* If $n\in\mathbb{N}$ and $k\in\mathbb{N}$ satisfy $k>n$, then
\begin{equation}
\genfrac{\{}{\}}{0pt}{0}{n}{k}=0
\label{darij1.eq.stir=0}
\tag{2}
\end{equation}
(since there is no set partition of the set $\left\{ 1,2,\ldots,n\right\} $
into more than $n$ nonempty subsets).


(Of course, you can use \eqref{darij1.eq.stir=} as a definition of $\genfrac{\{}{\}}{0pt}{0}{n}{k}$, but then these two facts are not obvious.)


For any $n\in\mathbb{N}$, we define a number $B\_{n}$ by
\begin{equation}
B\_{n}=\sum\_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{0}{n}{k}.
\end{equation}
This number $B\_{n}$ is the number of all set partitions of the set $\left\{
1,2,\ldots,n\right\} $ (since each addend $\genfrac{\{}{\}}{0pt}{0}{n}{k}$ in the sum counts those set partitions that have precisely $k$ parts). It is
known as a [Bell number](https://en.wikipedia.org/wiki/Bell_number).


The first Bell numbers are $B\_{0}=1$, $B\_{1}=1$, $B\_{2}=2$, $B\_{3}=5$ and
$B\_{4}=15$.


Now, for any $n\in\mathbb{N}$ and $p\in\mathbb{N}$, let me define the number
\begin{equation}
S\_{p}\left( n\right) =\sum\_{k=0}^{n}\dfrac{k^{p}}{k!}\sum\_{l=0}^{n-k}
\dfrac{\left( -1\right) ^{l}}{l!}.
\end{equation}


You are conjecturing that $S\_{1}\left( n\right) =1$ for all $n\geq1$, and
that $S\_{2}\left( n\right) =2$ for all $n\geq2$. These two equalities are
particular cases of the following fact:



> 
> **Theorem 1.** Let $n\in\mathbb{N}$ and $p\in\mathbb{N}$ satisfy $n\geq p$.
>  Then, $S\_{p}\left( n\right) =B\_{p}$.
> 
> 
> 


But we can prove something even better:



> 
> **Theorem 2.** Let $n\in\mathbb{N}$ and $p\in\mathbb{N}$. Then,
>  \begin{equation}
> S\_{p}\left( n\right) =\sum\_{k=0}^{n}
> \genfrac{\{}{\}}{0pt}{0}{p}{k}.
> \end{equation}
> 
> 
> 


*Proof of Theorem 2.* The definition of $S\_{p}\left( n\right) $ yields
\begin{align\*}
S\_{p}\left( n\right) & =\sum\_{k=0}^{n}\dfrac{k^{p}}{k!}\underbrace{\sum
\_{l=0}^{n-k}\dfrac{\left( -1\right) ^{l}}{l!}}\_{\substack{=\sum
\limits\_{j=k}^{n}\dfrac{\left( -1\right) ^{j-k}}{\left( j-k\right)
!}\\\text{(here, we have}\\\text{substituted }j-k\\\text{for }l\text{ in the
sum)}}}=\sum\_{k=0}^{n}\dfrac{k^{p}}{k!}\sum\_{j=k}^{n}\dfrac{\left( -1\right)
^{j-k}}{\left( j-k\right) !}\\
& =\underbrace{\sum\_{k=0}^{n}\sum\_{j=k}^{n}}\_{=\sum\_{j=0}^{n}\sum\_{k=0}^{j}
}\dfrac{k^{p}}{k!}\cdot\dfrac{\left( -1\right) ^{j-k}}{\left( j-k\right)
!}=\sum\_{j=0}^{n}\sum\_{k=0}^{j}\dfrac{k^{p}}{k!}\cdot\dfrac{\left( -1\right)
^{j-k}}{\left( j-k\right) !}\\
& =\sum\_{j=0}^{n}\underbrace{\sum\_{k=0}^{j}\dfrac{k^{p}}{k!}\cdot
\dfrac{\left( -1\right) ^{j-k}}{\left( j-k\right) !}}\_{\substack{=\sum
\limits\_{k=0}^{j}\dfrac{\left( j-k\right) ^{p}}{\left( j-k\right) !}
\cdot\dfrac{\left( -1\right) ^{k}}{k!}\\\text{(here, we have substituted
}j-k\\\text{for }k\text{ in the sum)}}}=\sum\_{j=0}^{n}\sum\_{k=0}
^{j}\underbrace{\dfrac{\left( j-k\right) ^{p}}{\left( j-k\right) !}
\cdot\dfrac{\left( -1\right) ^{k}}{k!}}\_{=\dfrac{1}{j!}\cdot\left(
-1\right) ^{k}\dfrac{j!}{k!\left( j-k\right) !}\left( j-k\right) ^{p}}\\
& =\sum\_{j=0}^{n}\sum\_{k=0}^{j}\dfrac{1}{j!}\cdot\left( -1\right)
^{k}\underbrace{\dfrac{j!}{k!\left( j-k\right) !}}\_{=\dbinom{j}{k}}\left(
j-k\right) ^{p}\\
& =\sum\_{j=0}^{n}\sum\_{k=0}^{j}\dfrac{1}{j!}\cdot\left( -1\right)
^{k}\dbinom{j}{k}\left( j-k\right) ^{p}=\sum\_{j=0}^{n}\dfrac{1}{j!}
\sum\_{k=0}^{j}\left( -1\right) ^{k}\dbinom{j}{k}\left( j-k\right) ^{p}\\
& =\sum\_{j=0}^{n}\dfrac{1}{j!}\sum\_{i=0}^{j}\left( -1\right) ^{i}\dbinom
{j}{i}\left( j-i\right) ^{p}
\end{align\*}
(here, we have renamed the summation index $k$ as $i$). Comparing this with
\begin{align\*}
& \sum\_{k=0}^{n}\underbrace{
\genfrac{\{}{\}}{0pt}{0}{p}{k}}\_{\substack{=\dfrac{1}{k!}\sum\limits\_{i=0}^{k}\left( -1\right) ^{i}
\dbinom{k}{i}\left( k-i\right) ^{p}\\\text{(by \eqref{darij1.eq.stir=},
applied to }p\text{ instead of }n\text{)}}}\\
& =\sum\_{k=0}^{n}\dfrac{1}{k!}\sum\limits\_{i=0}^{k}\left( -1\right)
^{i}\dbinom{k}{i}\left( k-i\right) ^{p}=\sum\_{j=0}^{n}\dfrac{1}{j!}
\sum\_{i=0}^{j}\left( -1\right) ^{i}\dbinom{j}{i}\left( j-i\right) ^{p}\\
& \qquad\left( \text{here, we have renamed the summation index }k\text{ as
}j\right) ,
\end{align\*}
we obtain $S\_{p}\left( n\right) =\sum\_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{0}{p}{k}$. This proves Theorem 2. $\blacksquare$


*Proof of Theorem 1.* For each $k\in\left\{ p+1,p+2,\ldots,n\right\} $, we
have $k\geq p+1>p$ and thus
\begin{equation}
\genfrac{\{}{\}}{0pt}{0}{p}{k}=0
\label{darij1.pf.t1.1}
\tag{3}
\end{equation}
(by \eqref{darij1.eq.stir=0}, applied to $p$ instead of $n$). Now, Theorem 2
yields
\begin{align\*}
S\_{p}\left( n\right) & =\sum\_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{0}{p}{k}\\
& =\sum\_{k=0}^{p}
\genfrac{\{}{\}}{0pt}{0}{p}{k}+\sum\_{k=p+1}^{n}\underbrace{
\genfrac{\{}{\}}{0pt}{0}{p}{k}}\_{\substack{=0\\\text{(by \eqref{darij1.pf.t1.1})}}}\\
& \qquad\left( \text{here, we have split the sum at }k=p\text{, since }0\leq
p\leq n\right) \\
& =\sum\_{k=0}^{p}
\genfrac{\{}{\}}{0pt}{0}{p}{k}+\underbrace{\sum\_{k=p+1}^{n}0}\_{=0}=\sum\_{k=0}^{p}
\genfrac{\{}{\}}{0pt}{0}{p}{k}=B\_{p}
\end{align\*}
(since the definition of $B\_{p}$ yields $B\_{p}=\sum\_{k=0}^{p}
\genfrac{\{}{\}}{0pt}{0}{p}{k}$). This proves Theorem 1. $\blacksquare$


