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
- Signed Stirling Number of the First Kind of n+1 with 1
- Stirling Number of the Second Kind of n+1 with 1
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(50)$