Unsigned Stirling Number of the First Kind of n+1 with 1

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Theorem

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

Then:

$\ds {n + 1 \brack 1} = n!$

where:

$\ds {n + 1 \brack 1}$ denotes an unsigned Stirling number of the first kind
$n!$ denotes $n$ factorial.


Proof

The proof proceeds by induction.


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

$\ds {n + 1 \brack 1} = n!$


Basis for the Induction

$\map P 0$ is the case:

\(\ds {1 \brack 1}\) \(=\) \(\ds \delta_{1 1}\) Unsigned Stirling Number of the First Kind of 1
\(\ds \) \(=\) \(\ds 1\) Definition of Kronecker Delta
\(\ds \) \(=\) \(\ds 0!\) Definition of Factorial


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

$\ds {k + 1 \brack 1} = k!$


from which it is to be shown that:

$\ds {k + 2 \brack 1} = \paren {k + 1}!$


Induction Step

This is the induction step:


\(\ds {k + 2 \brack 1}\) \(=\) \(\ds \paren {k + 1} {k + 1 \brack 1} + {k + 1 \brack 0}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds \paren {k + 1} {k + 1 \brack 1} + 0\) Unsigned Stirling Number of the First Kind of n+1 with 0
\(\ds \) \(=\) \(\ds \paren {k + 1} \times k!\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {k + 1}!\) Definition of Factorial


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


Therefore:

$\ds \forall n \in \Z_{\ge 0}: {n + 1 \brack 1} = n!$

$\blacksquare$


Also see


Sources