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 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:
 * $\displaystyle x^0 = \sum_k \left\{ {0 \atop k}\right\} x^{\underline k}$

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 $n = 0$.