Stirling Number of the Second Kind of Number with Greater/Proof 2

Theorem
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.

Proof
The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:


 * $\ds k > n \implies {n \brace k} = 0$

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


 * $\ds {0 \brace k} = \delta_{0 k}$

from Stirling Number of the Second Kind of 0.

So by definition of Kronecker delta:
 * $\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brace k} = 0$

and so $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\ds k > r \implies {r \brace k} = 0$

from which it is to be shown that:
 * $\ds k > r + 1 \implies {r + 1 \brace k} = 0$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$