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 $\map P n$ be the proposition:
 * the coefficients of the powers in the expression $\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$ are uniquely defined by $\ds {n \brack k} = {n - 1 \brack k - 1} + \paren {n - 1} {n - 1 \brack k}$

where $\ds {n \brack k} = \delta_{n k}$ 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:
 * $\ds {0 \brack k} = \delta_{0 k}$

Hence the result holds for $n = 0$.

Basis for the Induction
$\map P 1$ is the case:

We have:

Then:

Thus, in the expression:
 * $\ds x^{\underline k} = \sum_k \paren {-1}^{1 - k} {1 \brack k} x^1$

we have:
 * $\ds \paren {-1}^{1 - k} {1 \brack k} = 1$

and for all $k \in \Z$ where $k \ne 1$:
 * $\ds \paren {-1}^{1 - k} {1 \brack k} = 0$

That is:
 * $\ds \paren {-1}^{1 - k} {1 \brack k} = \delta_{1 k}$

Thus $\map P 1$ 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 $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

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

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

Induction Step
This is the induction step:

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

as required.

So $\map P r \implies \map P {r + 1}$ 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 $\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$ are uniquely defined by:
 * $\ds {n \brack k} = {n - 1 \brack k - 1} + \paren {n - 1} {n - 1 \brack k}$
 * where $\ds {n \brack k} = \delta_{n k}$ when $k = 0$ or $n = 0$.

Also see

 * Equivalence of Definitions of Signed Stirling Numbers of the First Kind
 * Equivalence of Definitions of Stirling Numbers of the Second Kind