Equivalence of Definitions of Stirling Numbers of the Second Kind

Theorem
The following definitions of the Stirling numbers of the second kind are equivalent:

Definition 2
in the sense that the coefficients of the falling factorial powers in the summand are uniquely defined by the given recurrence relation.

Proof
The Stirling numbers of the second kind

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * the coefficients of the falling factorial powers in the expression $\displaystyle x^n = \sum_k \left\{{n \atop k}\right\} x^{\underline k}$ are uniquely defined by $\displaystyle \left\{{n \atop k}\right\} =  \left\{{n - 1 \atop k - 1}\right\} + k \left\{{n - 1 \atop k}\right\}$

where $\displaystyle \left\{{n \atop k}\right\} = \delta_{n k}$ where $k = 0$ or $n = 0$.

First the case where $n = 0$ is attended to.

We have:

Thus, in the expression:
 * $\displaystyle x^0 = \sum_k \left\{ {n \atop k}\right\} x^{\underline k}$

we have:
 * $\displaystyle \left\{{0 \atop 0}\right\} = 1$

and for all $k \in \Z_{>0}$:
 * $\displaystyle \left\{{0 \atop k}\right\} = 0$

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

Hence the result holds for $n = 0$.

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

We have:

Then:

Thus, in the expression:
 * $\displaystyle x^1 = \sum_k \left\{ {1 \atop k}\right\} x^{\underline k}$

we have:
 * $\displaystyle \left\{{1 \atop 1}\right\} = 1$

and for all $k \in \Z$ where $k \ne 1$:
 * $\displaystyle \left\{{1 \atop k}\right\} = 0$

That is:
 * $\displaystyle \left\{{1 \atop k}\right\} = \delta_{1 k}$

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * The coefficients in the expression $\displaystyle x^r = \sum_k \left\{ {r \atop k}\right\} x^{\underline k}$ are uniquely defined by $\displaystyle \left\{{r \atop k}\right\} = \left\{{r - 1 \atop k - 1}\right\} + k \left\{{r - 1 \atop k}\right\}$

from which it is to be shown that:
 * The coefficients in the expression $\displaystyle x^{r + 1} = \sum_k \left\{ {r + 1 \atop k}\right\} x^{\underline k}$ are uniquely defined by $\displaystyle \left\{{r + 1 \atop k}\right\} = \left\{{r \atop k - 1}\right\} + k \left\{{r \atop k}\right\}$

Induction Step
This is the induction step:

Anticipating the expected result, we use $\displaystyle \left\{{r + 1 \atop k}\right\}$ to denote the coefficients of the $k$th falling factorial power in the expansion of $x^{r + 1}$.

Thus:

Thus the coefficients of the falling factorial powers are defined by the recurrence relation:
 * $\displaystyle \left\{{r + 1 \atop k}\right\} = \left\{ {r \atop {k - 1} }\right\} + k \left\{ {r \atop k}\right\}$

as required.

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

Therefore:
 * for all $n \in \Z_{\ge 0}$, the coefficients of the falling factorial powers in the expression $\displaystyle x^n = \sum_k \left\{ {n \atop k}\right\} x^{\underline k}$ are uniquely defined by:
 * $\displaystyle \left\{{n \atop k}\right\} = \left\{{n - 1 \atop k - 1}\right\} + k \left\{{n - 1 \atop k}\right\}$
 * where $\displaystyle \left\{{n \atop k}\right\} = \delta_{n k}$ where $k = 0$ or $n = 0$.