Summation of Products of n Numbers taken m at a time with Repetitions

Theorem
Let $a, b \in \Z$ be integers such that $b \ge a$.

Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.

Let $m \in \Z_{>0}$ be a (strictly) positive integer.

Let:

That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time, allowing for repetition.

For $r \in \Z_{> 0}$, let:
 * $S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$

Then:

Lemma 2
Then:

Thus, by definition of generating function:
 * $h_m = \ds \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$

Also presented as
When usually presented, set $U$ is usually defined to be either $\set {x_0, x_1, \ldots, x_n}$ or $\set {x_1, x_2, \ldots, x_n}$.

This leads to the summation $h_m$ to be defined either as:
 * $h_m = \ds \sum_{0 \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le n} \paren {\prod_{k \mathop = 1}^m x_{j_k} }$

or:
 * $h_m = \ds\sum_{1 \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le n} \paren {\prod_{k \mathop = 1}^m x_{j_k} }$

and for the summations $S_r$ to be defined as either $\ds \sum_{k \mathop = 0}^n {x_k}^r$ or $\ds \sum_{k \mathop = 1}^n {x_k}^r$ accordingly.

However, this may cause confusion when the indices are presented differently between separate invocations of this general result.

For example, in 's of $1997$:
 * in $\S 1.2.3$, $\displaystyle \sum_{k \mathop = 0}^n$ is used for the cases $m = 2$ and $n = 3$

while:
 * in $\S 1.2.9$, the same result has been taken over $\ds \sum_{k \mathop = 1}^n$

without explanatory comment.

Hence the decision has been made on to choose to define the set of indices over the general interval $\closedint a b$, to emphasise their essentially arbitrary nature.