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

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

Then:
 * $\ds {n + 1 \brace 1} = 1$

where $\ds {n + 1 \brace 1}$ 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:
 * $\ds {n + 1 \brace 1} = 1$

Basis for the Induction
$\map P 0$ is the case:

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:
 * $\ds {k + 1 \brace 1} = 1$

from which it is to be shown that:
 * $\ds {k + 2 \brace 1} = 1$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$

Also see

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


 * Particular Values of Stirling Numbers of the Second Kind