Combinatorial proof of $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!$, using inclusion-exclusion If $l$ and $n$ are any positive integers, is there a  proof of the identity

$$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!\;$$

which uses the Inclusion-Exclusion Principle?
(If necessary, restrict to the case where $l\ge n$.)

This question is closely related to Expressing a factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$?
and also Proof of the summation $n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$?
 A: Assume that $\ell\ge n$. We want to count the injections from $[n]$ to $[\ell]$ whose range is $[n]$. For each $k\in[n]$ let $A_k$ be the set of functions from $[n]$ to $[\ell]\setminus\{k\}$. It’s not hard to see that for any non-empty $I\subseteq[n]$ we have
$$\left|\,\bigcap_{k\in I}A_k\,\right|=(\ell-|I|)^n\;,$$
so by the inclusion-exclusion principle we have
$$\begin{align*}\left|\,\bigcup_{k=1}^nA_k\,\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|-1}(\ell-|I|)^n\\
&=\sum_{k=1}^n\binom{n}k(-1)^{k-1}(\ell-k)^n\;.
\end{align*}$$
This is the number of functions from $[n]$ to $[\ell]$ that miss at least one element of $[n]$, so we want the size of the complementary set, which is
$$\begin{align*}
\ell^n-\sum_{k=1}^n\binom{n}k(-1)^{k-1}(\ell-k)^n&=(-1)^0\binom{n}0(\ell-0)^n+\sum_{k=1}^n(-1)^k\binom{n}k(\ell-k)^n\\
&=\sum_{k=0}^n(-1)^k\binom{n}k(\ell-k)^n\;.
\end{align*}$$
Of course there are $n!$ injections from $[n]$ to $[\ell]$ with range $[n]$, so 
$$\sum_{k=0}^n(-1)^k\binom{n}k(\ell-k)^n=n!\tag{1}$$
for $\ell\ge n$. 
Let
$$p(x)=n!-\sum_{k=0}^n(-1)^k\binom{n}k(x-k)^n\;;$$
$p(x)$ is a polynomial in $x$ of degree $n$, and every integer $\ell\ge n$ is a zero of $p(x)$, so $p(x)$ must be constant, and therefore
$$\sum_{k=0}^n(-1)^k\binom{n}k(x-k)^n=n!$$
for all $x$: $x$ need not even be an integer.
A: Let $A=\{1,\cdots,n\}$ and $B=\{1,\cdots,l\}$ where $l\ge n$, and let $S$ be the set of functions from $A$ to $B$.  
If $E_i$ is the set of functions in $S$ which do not have the value $i$ for $1\le i\le n$, 
then $\overline{E_1}\cap\cdots\cap\overline{E_n}$ is the set of functions from $A$ to $B$ which have $A$ as their range.
Using Inclusion-Exclusion,
$\displaystyle|\overline{E_1}\cap\cdots\cap\overline{E_n}|=|S|-\sum_{i}|E_i|+\sum_{i<j}|E_i\cap E_j|-\sum_{i<j<k}|E_i\cap E_j\cap E_k|+\cdots$ 
so $\;\displaystyle n!=\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^{n}$.
A: NOTE: I am providing an answer here as the original post I had wrote it in (see here) was made a duplicate of this post. As such, I reference this post (as in the exact page you the reader are on right now) in my answer. I believe my answer is sufficiently different to warrant this however as I expand on the question and provide an interesting approach using analysis. Here is my answer to the other question in full

We will prove something stronger. Namely, that
$$f_k(x)=\sum_{n=0}^k\frac{(k+x-n)^k(-1)^n}{(k-n)!n!}=1$$
for any $x\in\mathbb{R}$. However, first we will provide a lemma that will be used later in the proof. That is that identity
$$\sum_{n=0}^k\binom{k}{n}n^j(-1)^n=0\text{ for }0\leq j<k$$
From here, an explicit formula for Stirling Numbers of the Second Kind is
$$S2(j,k)=\frac{1}{k!}\sum_{n=0}^k(-1)^{k-n}\binom{k}{n}n^j$$
Then
$$\sum_{n=0}^k\binom{k}{n}n^j(-1)^n=k!(-1)^kS2(j,k)$$
However, $S2(j,k)$ is defined as the number of ways to arrange $j$ elements into $k$ non-empty sets. The key here being non-empty. That is, if $j<k$ then it is impossible to arrange the $j$ elements into $k$ non-empty sets so $S2(j,k)=0$. That is, if $j<k$ then
$$\sum_{n=0}^k\binom{k}{n}n^j(-1)^n=0$$
Having proved our lemma, we continue onto the main proof. Now, consider the form of $f_k(x)$. Obviously, $f_k(x)$ is a polynomial. That is, it is an entire function. We are justified then in taking its Maclaurin Series. To this end, we need to calculate $f_k^{(m)}(0)$ for $m\geq 1$. Now, since $f_k(x)$ is of degree at most $k$, $f_k^{(m)}(0)=0$ for $m\geq k+1$. Consider the derivatives of $f_k(x)$:
$$f_k^{(0)}(x)=\sum_{n=0}^k\frac{(-1)^n (k-n+x)^k}{n! (k-n)!}$$
$$f_k^{(1)}(x)=\sum_{n=0}^k\frac{k (-1)^n (k-n+x)^{k-1}}{n! (k-n)!}$$
$$f_k^{(2)}(x)=\sum_{n=0}^k\frac{(k-1) k (-1)^n (k-n+x)^{k-2}}{n! (k-n)!}$$
$$\vdots$$
$$f_k^{(m)}(x)=\sum_{n=0}^k\frac{k! (-1)^n (k-n+x)^{k-m}}{n! (k-n)!(k-m)!}$$
for $0\leq m\leq k$. Thus,
$$f_k^{(m)}(0)=\sum_{n=0}^k\frac{k! (-1)^n (k-n)^{k-m}}{n! (k-n)!(k-m)!}$$
$$=\frac{1}{(k-m)!}\sum_{n=0}^k\frac{k! (-1)^n (k-n)^{k-m}}{n! (k-n)!}$$
Expanding $(k-n)^{k-m}$ in its Binomial Series gives us
$$=\frac{1}{(k-m)!}\sum_{n=0}^k\left[\binom{k}{n} \sum_{j=0}^{k-m}\binom{k-m}{j}k^{k-m-j}n^j(-1)^{n+j}\right]$$
Since these are finite sums, we can switch the order of summation to get
$$=\frac{1}{(k-m)!}\sum_{j=0}^{k-m}\left[ \binom{k-m}{j}k^{k-m-j}(-1)^n\sum_{n=0}^{k}\binom{k}{n}n^j(-1)^{j}\right]$$
But for $m\geq 1$, $j$ will always be less than $k$. This is precisely the lemma we proved above. Thus, for $m\geq 1$
$$=\frac{1}{(k-m)!}\sum_{j=0}^{k-m}\left[ \binom{k-m}{j}k^{k-m-j}(-1)^n\cdot 0\right]=0$$
and hence
$$f_k^{m}(0)=0$$
Since every coefficient in the Maclaurin Series except for the $m=0$ is zero, we conclude $f_k(x)$ is a constant function. That is
$$f_k(x)=a_k$$
where $a_k$ could change depending on $k$. We seek to show that $a_k=1$ for all $k$. In fact, this is easily done as we can simply calculate $f_k(0)$:
$$f_k(0)=\sum_{n=0}^k\frac{(k-n)^k(-1)^n}{(k-n)!n!}=a_k$$
(this is your original question). In fact, these terms can be rearrange to 
$$a_k=\sum_{n=0}^k\frac{(k-n)^k(-1)^n}{(k-n)!n!}=\frac{1}{k!}\sum_{n=0}^k\frac{k!(k-n)^k(-1)^n}{(k-n)!n!}$$
$$k!a_k=\sum_{n=0}^k\binom{k}{n}(k-n)^k(-1)^n$$
However, as was pointed out by user @Angela_Richardson, this exact problem can be found here already. In the link provided, simply swap $n$ and $k$, and set $l=k$ to get
$$k!a_k=\sum_{n=0}^k\binom{k}{n}(k-n)^k(-1)^n=k!$$
$$a_k=1$$
Thus, your original question is answered in the affirmative as well as
$$f_k(x)=\sum_{n=0}^k\frac{(k+x-n)^k(-1)^n}{(k-n)!n!}=1$$
for all $x\in\mathbb{R}$.
