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How the two following formulas can be proved (algebraically preferred)? $$\sum_{n=k}^{∞}S\left(n,k\right)\ \frac{x^{n}}{n!}=\frac{1}{k!}\left(e^{x}-1\right)^{k}$$ $$x^{n}=\sum_{m=0}^{n}S\left(n,m\right)\left(x\right)_{m}$$ where $S\left(n,m\right)$ is the Stirling number of the second kind and $\left(x\right)_{m}$ is falling factorial. any Hint or full proof is highly appreciated (since I'm new to this numbers and their relations so a full proof is better).

Absurd
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  • What is your starting point? Like what is your definition of Stirling number. The combinatorial description? Or the connecting coefficients(mainly your second equation) – Phicar Nov 01 '19 at 13:55
  • my definition is " the number of ways to partition a set of $m$ objects into $k$ non-empty subsets is called Stirling number of the second kind", Also I just saw these formulas at Wolfram Mathworld and did not have any idea to prove them. – Absurd Nov 01 '19 at 14:00
  • For the first one, do you know something about exponential map of combinatorial classes? .For the second one, take a function $f$ from $[n]$ to $[m].$ Consider then the image $Im(f)$ and suppose $1\leq|Im(f)={m_1,\cdots ,m_k}|=k\leq m$ This induces a partition of $[n]$ as $\pi ={f^{-1}(m_1),\cdots f^{-1}(m_k)}$ and by taking the quotient $[n]/\pi$ we have an injective function. Can you finish? If not, i will answer fully. – Phicar Nov 01 '19 at 16:16

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If you are familiar with combinatorial classes the first equation represents a set of $k$ elements, each of these themselves a set of labeled elements, with a non-zero number of elements (here we are distributing $n$ distinguishable balls into $k$ indistinguishable boxes with no box being empty):

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}_{=k} (\textsc{SET}_{\ge 1}(\mathcal{Z})).$$

The exponential generating function for $\textsc{SET}_{=k}(\mathcal{Z})$ is $$\frac{z^k}{k!}.$$ This yields for set partitions into $k$ sets

$$\frac{1}{k!} \left(\sum_{q=1}^\infty \frac{z^q}{q!}\right)^k = \frac{1}{k!} (\exp(z)-1)^k.$$

This is because in a combinatorial class constructed by distributing a repertoire of source objects into a row of $k$ slots with a group $G$ permuting the slots and creating equivalence classes, the EGF of the class is given by

$$\frac{S(z)^k}{|G|}.$$

where $S(z)$ is the EGF of the source objects. For sets $G$ is the symmetric group and $|G|=k!.$ These combinatorial classes may be nested. Another example are cycles with $|G|=k$ where we get the generating function

$$\sum_{q\ge 1} \frac{z^q}{q} = \log\frac{1}{1-z}$$

for the class of cycles. Of course permutations are sets of cycles and we have

$$n! [z^n] \exp \log\frac{1}{1-z} = n! [z^n] \frac{1}{1-z} = n!.$$

For the second equation we have the following claim where $x$ is a positive integer:

$$x^n = \sum_{m=0}^n {n\brace m} x^{\underline m} = \sum_{m=1}^n {x\choose m} m! {n\brace m}.$$

Suppose we throw $n$ different balls into $x$ different boxes, there are $x^n$ ways of doing this. On the other hand we may classify every distribution of balls obtained in this way by the number $m$ of boxes that were not empty. To get this kind of distribution we choose the $m$ boxes in ${x\choose m}$ ways and partition the $n$ balls into $m$ sets in ${n\brace m}$ ways. These $m$ sets can be matched to the chosen $m$ boxes in $m!$ ways and every such configuration constitutes a distribution of the balls, and we have equality.

Given that LHS and RHS may be viewed as polynomials in $x$ and they are equal at all positive integer values, they are the same polyonmial.

Marko Riedel
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  • Thank you so much for your great answe (+1)but this is so advanced for someone like me and that's because the lack of my knowledge ,it would be highly appreciated if you tell me some good references which I can learn what you used in your answer. – Absurd Nov 01 '19 at 18:04
  • The notation for the exponential generating function and the set operator (class) is from Analytic Combinatorics by Flajolet and Sedgewick. With the second part it mostly seems to be in the public domain. – Marko Riedel Nov 01 '19 at 18:47
  • @MarkoRiedel: Very nice and thoughtful answer. (+1) – Markus Scheuer Nov 02 '19 at 11:25
  • Thank you for the kind remark. – Marko Riedel Nov 02 '19 at 16:00