Proving $\sum^k_{n=0}\frac{(k-n)^k(-1)^n}{(k-n)!n!}=1$ Please look at the following relation.
$$\forall k \in \mathbb{N}, \ \ \ \ \ \sum^k_{n=0}\frac{(k-n)^k(-1)^n}{(k-n)!n!}=1$$
I don't know a proof of this relation. 
I tried mathematical induction. But it didn't work.
 A: 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
$$=\frac{1}{(k-m)!}\sum_{j=0}^{k-m}\left[ \binom{k-m}{j}k^{k-m-j}(-1)^j\sum_{n=0}^{k}\binom{k}{n}n^j(-1)^{n}\right]$$
But for $m\geq 1$, $j$ will always be less than $k$. This is precisely the lemma we proved above. Thus, for $m\geq 1$
$$=\frac{1}{(k-m)!}\sum_{j=0}^{k-m}\left[ \binom{k-m}{j}k^{k-m-j}(-1)^j\cdot 0\right]=0$$
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 $x\in\mathbb{R}$.
