Stirling Number of the Second Kind of n+1 with 1

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Theorem

Let $n \in \Z_{\ge 0}$.

Then:

$\displaystyle {n + 1 \brace 1} = 1$

where $\displaystyle {n + 1 \brace 1}$ denotes a Stirling number of the second kind.


Proof

The proof proceeds by induction.


For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\displaystyle {n + 1 \brace 1} = 1$


Basis for the Induction

$\map P 0$ is the case:

\(\ds {1 \brace 1}\) \(=\) \(\ds \delta_{1 1}\) Stirling Number of the Second Kind of 1
\(\ds \) \(=\) \(\ds 1\) Definition of Kronecker Delta


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\displaystyle {k + 1 \brace 1} = 1$


from which it is to be shown that:

$\displaystyle {k + 2 \brace 1} = 1$


Induction Step

This is the induction step:


\(\ds {k + 2 \brace 1}\) \(=\) \(\ds 1 \times {k + 1 \brace 1} + {k + 2 \brace 0}\) Definition of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds {k + 1 \brace 1} + 0\) Stirling Number of the Second Kind of n+1 with 0
\(\ds \) \(=\) \(\ds 1\) Induction Hypothesis


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


Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$

$\blacksquare$


Also see


Sources