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 strong induction.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:


 * $\displaystyle k > n \implies \left\{ {n \atop k}\right\} = 0$

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


 * $\displaystyle \left\{ {0 \atop k}\right\} = \delta_{0 k}$

from Stirling Number of the Second Kind of 0.

So by definition of Kronecker delta:
 * $\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle \left\{ {0 \atop k}\right\} = 0$

and so $P \left({0}\right)$ is seen to hold.

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

Thus by definition of Kronecker delta:
 * $\forall k \in \Z_{\ge 0}: k > 1 \implies \displaystyle \left\{ {1 \atop k}\right\} = 0$

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\displaystyle k > r \implies \left\{ {r \atop k}\right\} = 0$

from which it is to be shown that:
 * $\displaystyle k > r + 1 \implies \left\{ {r + 1 \atop k}\right\} = 0$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies \left\{ {n \atop k}\right\} = 0$