Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 2

Proof
From Summation to n of Summation to Index:


 * $\displaystyle \sum_{1 \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le n} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$

where:
 * $S_j = \displaystyle \sum_{k \mathop = a}^b {x_k}^j$ for $j \in \Z_{\ge 0}$

Setting $m = 2$:

The result follows by rearranging indices.