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

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

Then:
 * $\displaystyle \left\{ {n + 1 \atop 2}\right\} = 2^n - 1$

where $\displaystyle \left\{ {n + 1 \atop 2}\right\}$ denotes a Stirling number of the second kind.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \left\{ {n + 1 \atop 2}\right\} = 2^n - 1$

$P \left({0}\right)$ is the case:

So the result holds for $P \left({0}\right)$.

Basis for the Induction
$P \left({1}\right)$ is the case:

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \left\{ {k + 1 \atop 2}\right\} = 2^k - 1$

from which it is to be shown that:
 * $\displaystyle \left\{ {k + 2 \atop 2}\right\} = 2^{k + 1} - 1$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \left\{ {n + 1 \atop 2}\right\} = 2^n - 1$

Also see

 * Particular Values of Stirling Numbers of the Second Kind