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

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

Then:
 * $\displaystyle {n \brace n - 1} = \binom n 2$

where:
 * $\displaystyle {n \brace n - 1}$ 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 $\map P n$ be the proposition:
 * $\displaystyle {n \brace n - 1} = \binom n 2$

Basis for the Induction
$\map P 1$ 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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\displaystyle {k \brace k - 1} = \binom k 2$

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

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

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