Stirling Number of the Second Kind of Number with Self

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Theorem

$\displaystyle \left\{ {n \atop n}\right\} = 1$

where $\displaystyle \left\{ {n \atop n}\right\}$ denotes a Stirling number of the second kind.


Proof

The proof proceeds by induction.


For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \left\{ {n \atop n}\right\} = 1$


Basis for the Induction

$P \left({0}\right)$ is the case:

\(\displaystyle \left\{ {0 \atop 0}\right\}\) \(=\) \(\displaystyle \delta_{0 0}\) Stirling Number of the Second Kind of 0
\(\displaystyle \) \(=\) \(\displaystyle 1\) Definition of Kronecker Delta


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is the induction hypothesis:

$\displaystyle \left\{ {k \atop k}\right\} = 1$


from which it is to be shown that:

$\displaystyle \left\{ {k + 1 \atop k + 1}\right\} = 1$


Induction Step

This is the induction step:


\(\displaystyle \left\{ { {k + 1} \atop {k + 1} }\right\}\) \(=\) \(\displaystyle \left({k + 1}\right) \left\{ {k \atop {k + 1} }\right\} + \left\{ {k \atop k}\right\}\) Definition of Stirling Numbers of the Second Kind
\(\displaystyle \) \(=\) \(\displaystyle \left({k + 1}\right) \times 0 + \left\{ {k \atop k}\right\}\) Stirling Number of Number with Greater
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {k \atop k}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\) Induction Hypothesis


So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: \left\{ {n \atop n}\right\} = 1$

$\blacksquare$


Also see


Sources