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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.

Bruno B
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RTF
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    Does this answer your question? sum of alternating binomial There are many duplicates. – Anne Bauval Aug 03 '23 at 06:43
  • It's easy to see that $$ \left(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \right)^n = \left( \frac{e^x-1}{x} \right)^n$$ which is not $1$. – awkward Aug 03 '23 at 13:15
  • @awkward If you check out the link, the answerer ends up concluding that the coefficient of $x^n$ is both $\sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^n}{n!} = 1$ and $\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots\right)^n \neq 1$. How does one make sense of that? – RTF Aug 03 '23 at 14:11
  • @AnneBauval Not really. My question is about an identity that contains the alternating binomial coefficients, but not exactly that identity itself. Moreover, I am interested in the specific argument I linked. – RTF Aug 03 '23 at 14:14
  • The claim in the linked-to solution is not that $ \left(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \right)^n = 1$ but that the coefficient of $x^n$ when the expression is expanded is $1$. – awkward Aug 03 '23 at 14:36
  • @akward "To finish up, we can go back to our original expression which was $\cdots =x^n\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots\right)^n$." I may be missing something obvious, but if we have $x^n \times \left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots\right)^n$, how is $\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots\right)^n$ not its coefficient and thus equal to any other coefficient previously derived for it? – RTF Aug 03 '23 at 15:04
  • @awkward (Sorry, mistagged you in previous reply). – RTF Aug 03 '23 at 15:26
  • The duplicate I selected (among many others to which I linked) gives several proofs of the identity you are asking for. With a very small effort, you will realize that it actually is exactly "that identity itself". – Anne Bauval Aug 03 '23 at 17:59

2 Answers2

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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 $1$.

Jim Ferry
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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.

NoName
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