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:
Let:
{{begin-eqn}}
{{eqn | n = a
      | l = A
      | o = :=
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k
      | c = 
}}
{{eqn | r = \sum_{a \mathop \le i \mathop \le j \mathop \le k \mathop \le b} x_i x_j x_k
      | c = 
}}
{{eqn | n = b
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \sum_{k \mathop = j}^b x_i x_j x_k
      | c = 
}}
{{eqn | n = c
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \sum_{k \mathop = i}^j x_i x_j x_k
      | c = 
}}
{{eqn | n = d
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = i}^j x_i x_j x_k
      | c = 
}}
{{eqn | n = e
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = i}^b x_i x_j x_k
      | c = 
}}
{{end-eqn}}
Also, let:
{{begin-eqn}}
{{eqn | l = S_1
      | o = :=
      | r = \sum_{i \mathop = a}^b x_i
      | c = 
}}
{{eqn | l = S_2
      | o = :=
      | r = \sum_{i \mathop = a}^b {x_i}^2
      | c = 
}}
{{eqn | l = S_3
      | o = :=
      | r = \sum_{i \mathop = a}^b {x_i}^3
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = 2 A
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \sum_{k \mathop = j}^b x_i x_j x_k + \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \sum_{k \mathop = i}^j x_i x_j x_k
      | c = $(b) + (c)$
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \left({\sum_{k \mathop = j}^b x_i x_j x_k + \sum_{k \mathop = i}^j x_i x_j x_k}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \left({\sum_{k \mathop = i}^b x_i x_j x_k + x_i x_j x_j}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = i}^b x_j}\right)^2 + \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = i}^b {x_j}^2}\right)
      | c = 
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = A_1
      | o = :=
      | r = \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = i}^b x_j}\right)^2
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b x_i \sum_{j \mathop = i}^b x_j \sum_{k \mathop = i}^b x_k
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \sum_{k \mathop = i}^b x_i x_j x_k
      | c = 
}}
{{eqn | l = A_3
      | o = :=
      | r = \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = i}^b {x_j}^2}\right)
      | c = 
}}
{{end-eqn}}
as calculated above.
Thus:
:$(1): \quad 2 A = A_1 + A_3$
Similarly:
{{begin-eqn}}
{{eqn | l = 2 A
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k + \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = j}^i x_i x_j x_k
      | c = $(a) + (d)$
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \left({\sum_{k \mathop = a}^j x_i x_j x_k + \sum_{k \mathop = j}^i x_i x_j x_k}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \left({\sum_{k \mathop = a}^i x_i x_j x_k + x_i x_j x_j}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = a}^i x_j}\right)^2 + \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = a}^i {x_j}^2}\right)
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = A_1 + A
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b \sum_{k \mathop = i}^b x_i x_j x_k + \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = i}^b x_i x_j x_k
      | c = using $(e)$
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{k \mathop = i}^b \left({\sum_{j \mathop = i}^b x_i x_j x_k + \sum_{j \mathop = a}^i x_i x_j x_k}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{k \mathop = i}^b \left({\sum_{j \mathop = a}^b x_i x_j x_k + {x_i}^2 x_k}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b x_i \sum_{j \mathop = a}^b \sum_{k \mathop = j}^b x_j x_k + \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b {x_i}^2 x_j
      | c = 
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = A_2
      | o = :=
      | r = \sum_{i \mathop = a}^b x_i \sum_{j \mathop = a}^b \sum_{k \mathop = j}^b x_j x_k
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b \sum_{k \mathop = j}^b x_i x_j x_k
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b \sum_{k \mathop = a}^j x_i x_j x_k
      | c = 
}}
{{eqn | l = A_4
      | o = :=
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b {x_i}^2 x_j
      | c = 
}}
{{end-eqn}}
as calculated above.
Thus:
:$(2): \quad A_1 + A = A_2 + A_4$
Then:
{{begin-eqn}}
{{eqn | l = 2 A_2
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b \sum_{k \mathop = j}^b x_i x_j x_k + \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b \sum_{k \mathop = a}^j x_i x_j x_k
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b \left({\sum_{k \mathop = j}^b x_i x_j x_k + \sum_{k \mathop = a}^j x_i x_j x_k}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b \left({\sum_{k \mathop = a}^b x_i x_j x_k + x_i {x_j}^2}\right)
      | c = 
}}
{{eqn | r = \left({\sum_{i \mathop = a}^b x_i}\right)^3 + \sum_{i \mathop = a}^b x_i \sum_{i \mathop = a}^b {x_i}^2
      | c = 
}}
{{eqn | n = 3
      | r = {S_1}^3 + S_1 S_2
      | c = 
}}
{{end-eqn}}
Now we have that:
{{begin-eqn}}
{{eqn | l = A_3
      | r = \sum_{i \mathop = a}^b x_i \left({\sum_{j \mathop = i}^b {x_j}^2}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b {x_i}^2 \left({\sum_{j \mathop = a}^i x_j}\right)
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = A_3 + A_4
      | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i {x_i}^2 x_j + \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b {x_i}^2 x_j
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b \left({\sum_{j \mathop = a}^b {x_i}^2 x_j + {x_i}^3}\right)
      | c = 
}}
{{eqn | r = \sum_{i \mathop = a}^b {x_i}^2 \sum_{i \mathop = a}^b x_i + \sum_{i \mathop = a}^b {x_i}^3
      | c = 
}}
{{eqn | n = 4
      | r = S_2 S_1 + S_3
      | c = 
}}
{{end-eqn}}
Finally:
{{begin-eqn}}
{{eqn | l = 2 A
      | r = A_1 + A_3
      | c = from $(1)$
}}
{{eqn | l = A + A_1
      | r = A_2 + A_4
      | c = from $(2)$
}}
{{eqn | l = 2 A_2
      | r = {S_1}^3 + S_1 S_2
      | c = from $(3)$
}}
{{eqn | l = A_3 + A_4
      | r = S_1 S_2 + S_3
      | c = from $(4)$
}}
{{eqn | ll= \leadsto
      | l = 3 A + A_1
      | r = A_1 + A_2 + A_3 + A_4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 A
      | r = A_2 + A_3 + A_4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 6 A
      | r = 2 A_2 + 2 \left({A_3 + A_4}\right)
      | c = 
}}
{{eqn | r = {S_1}^3 + S_1 S_2 + 2 S_1 S_2 + 2 S_3
      | c = 
}}
{{eqn | r = {S_1}^3 + 3 S_1 S_2 + 2 S_3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = \frac { {S_1}^3} 6 + \frac {S_1 S_2} 2 + \frac {S_3} 3
      | c = 
}}
{{end-eqn}}
{{qed}}