Summation over Lower Index of Unsigned Stirling Numbers of the First Kind

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Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

$\ds \sum_k {n \brack k} = n!$

where:

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


Proof

The proof proceeds by induction on $n$.


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

$\ds \sum_k {n \brack k} = n!$


$\map P 0$ is the case:

\(\ds \sum_k {0 \brack k}\) \(=\) \(\ds \sum_k \delta_{0 k}\) Unsigned Stirling Number of the First Kind of 0
\(\ds \) \(=\) \(\ds 1\) all terms vanish but for $k = 0$
\(\ds \) \(=\) \(\ds 0!\) Definition of Factorial

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds \sum_k {1 \brack k}\) \(=\) \(\ds \sum_k \delta_{1 k}\) Unsigned Stirling Number of the First Kind of 1
\(\ds \) \(=\) \(\ds 1\) all terms vanish but for $k = 1$
\(\ds \) \(=\) \(\ds 1!\) Definition of Factorial

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

$\ds \sum_k {m \brack k} = m!$


from which it is to be shown that:

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


Induction Step

This is the induction step:


\(\ds \sum_k {m + 1 \brack k}\) \(=\) \(\ds \sum_k \paren {m {m \brack k} + {m \brack k - 1} }\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds m \sum_k {m \brack k} + \sum_k {m \brack k - 1}\)
\(\ds \) \(=\) \(\ds m \sum_k {m \brack k} + \sum_k {m \brack k}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \paren {m + 1} \sum_k {m \brack k}\)
\(\ds \) \(=\) \(\ds \paren {m + 1} m!\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {m + 1}!\) Definition of Factorial


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


Therefore:

$\ds \forall n \in \Z_{\ge 0}: \sum_k {n \brack k} = n!$

$\blacksquare$


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