Stirling Number of the Second Kind of n+1 with 2

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

Then:
 * $\ds {n + 1 \brace 2} = 2^n - 1$

where $\ds {n + 1 \brace 2}$ 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:
 * $\ds {n + 1 \brace 2} = 2^n - 1$

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

So the result holds for $\map P 0$.

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:
 * $\ds {k + 1 \brace 2} = 2^k - 1$

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

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:
 * $\ds \forall n \in \Z_{\ge 0}: {n + 1 \brace 2} = 2^n - 1$

Also see

 * Particular Values of Stirling Numbers of the Second Kind