About the value of $\sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^n}{n!}$

Question

Is

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

a common identity?

After extensive googling, I am unable to find anything on it specifically.

It came up as part of an answer to a previous binomial question of mine: Binomial proof of $n! = n^r - n(n - 1)^r + \frac{n(n-1)}{2!}\left(n -2 \right)^r - \dots$ when $r = n$. I follow the entire argument of the accepted answer, but I don't see why $\sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^n}{n!} = 1$. I'd prefer clarification within the context of the linked question, but combinatorial proofs are nice too.

Also, based on equating coefficients, the answer in the linked question implies that $$ \left(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \right)^n = 1$$

as well. I'm unfamiliar with this identity too, although it sort of looks like a Taylor series for...$1$??

Thank you for any assistance.

Answer 1

Reindexing your sum by replacing $k$ with $n-k$, we may observe that it is a special case of a Stirling number of the second kind, which is defined as$$ {n \brace j} = \frac{1}{j!} \sum_{k=0}^j (-1)^{j-k} \binom{j}{k}k^n. $$ This is the number of ways to partition a set of $n$ labelled objects into $j$ nonempty, unlabelled subsets. Your sum is ${n \brace n}$ which equals $1$.

Answer 2

Shift the index $\displaystyle \sum_{k=a}^{b}f(k) = \sum_{k=a}^{b}f(a+b-k)$ and recall $\displaystyle \binom {n}{k} = \binom {n}{n-k}$:

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

By this result this is equal to $\displaystyle \frac{(-1)^n}{n!}\cdot (-1)^n n! = 1.$

I believe this is called Tepper's identity.