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

Theorem
Let $n \in \Z_{> 0}$ be an integer greater than $0$.

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

where:
 * $\displaystyle \left\{ {n \atop n - 1}\right\}$ denotes a Stirling number of the second kind
 * $\dbinom n 2$ denotes a binomial coefficient.

Proof
The proof proceeds by induction.

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

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

Basis for the Induction
$P \left({2}\right)$ 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:
 * $\displaystyle \left\{ {k \atop k - 1}\right\} = \binom k 2$

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

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:
 * $\displaystyle \forall n \in \Z_{\ge 1}: \left\{ {n \atop n - 1}\right\} = \binom n 2$

Also see

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


 * Particular Values of Stirling Numbers of the Second Kind