Equivalence of Definitions of Unsigned Stirling Numbers of the First Kind

Definition 2
in the sense that the coefficients of the 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 powers in the expression $\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$ are uniquely defined by $\displaystyle \left[{n \atop k}\right] = \left[{n - 1 \atop k - 1}\right] + \left({n - 1}\right) \left[{n - 1 \atop k}\right]$

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.

From Unsigned Stirling Number of the First Kind of 0 we have:
 * $\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$

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:
 * $\displaystyle x^{\underline k} = \sum_k \left({-1}\right)^{1 - k} \left[{1 \atop k}\right] x^1$

we have:
 * $\displaystyle \left({-1}\right)^{1 - k} \left[{1 \atop k}\right] = 1$

and for all $k \in \Z$ where $k \ne 1$:
 * $\displaystyle \left({-1}\right)^{1 - k} \left[{1 \atop k}\right] = 0$

That is:
 * $\displaystyle \left({-1}\right)^{1 - k} \left[{1 \atop k}\right] = \delta_{1 k}$

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 $\displaystyle x^{\underline r} = \sum_k \left({-1}\right)^{r - k} \left[{r \atop k}\right] x^k$ are uniquely defined by $\displaystyle \left[{r \atop k}\right] = \left[{r - 1 \atop k - 1}\right] + \left({r - 1}\right) \left[{r - 1 \atop k}\right]$

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

Induction Step
This is the induction step:

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

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 powers in the expression $\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$ are uniquely defined by:
 * $\displaystyle \left[{n \atop k}\right] = \left[{n - 1 \atop k - 1}\right] + \left({n - 1}\right) \left[{n - 1 \atop k}\right]$
 * where $\displaystyle \left[{n \atop k}\right] = \delta_{n k}$ when $k = 0$ or $n = 0$.

Also see

 * Equivalence of Definitions of Stirling Numbers of the Second Kind