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

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:
{{begin-eqn}}
{{eqn | l = h_m
      | r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \paren {\prod_{k \mathop = 1}^m x_{j_k} }
      | c = 
}}
{{eqn | r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = h_m
      | r = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \paren {\prod_{j \mathop = 1}^m \dfrac { {S_j}^{k_j} } {j^{k_j} k_j !} }
      | c = 
}}
{{eqn | r = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \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 !}
      | c = 
}}
{{end-eqn}}

Proof:
We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$
From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$
where:
:$S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$.
Setting $m = 3$:
{{begin-eqn}}
{{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}
      | r = \sum_{\substack {k_1, k_2, k_3 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + 3 k_3 \mathop = 3} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !}
      | c = 
}}
{{end-eqn}}
We need to find all sets of $k_1, k_2, k_3 \in \Z_{\ge 0}$ such that:
:$k_1 + 2 k_2 + 3 k_3 = 3$
Thus $\tuple {k_1, k_2, k_3}$ can be:
:$\tuple {3, 0, 0}$
:$\tuple {1, 1, 0}$
:$\tuple {0, 0, 1}$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}
      | r = \dfrac { {S_1}^3 {S_2}^0 {S_3}^0} {\paren {1^3 \times 3!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} } + \dfrac { {S_1}^1 {S_2}^1 {S_3}^0} {\paren {1^1 \times 1!} \paren {2^1 \times 1!} \paren {3^0 \times 0!} } + \dfrac { {S_1}^0 {S_2}^0 {S_3}^1} {\paren {1^0 \times 0!} \paren {2^0 \times 0!} \paren {3^1 \times 1!} }
      | c = 
}}
{{eqn | r = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3
      | c = 
}}
{{end-eqn}}
{{qed}}