Recursion Property of Elementary Symmetric Function/Proof 1

Proof
Case $m = 1$ holds because $e_0$ is $1$ and $e_1$ is the sum of the elements.

Assume $2 \le m \le n$.

Define four sets:


 * $A = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n + 1}$


 * $B = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_{m - 1} \le n, p_m = n + 1}$


 * $C = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n}$


 * $D = \set {\set {p_1, \ldots, p_{m - 1} } : 1 \le p_1 < \cdots < p_{m - 1} \le n}$

Then $A = B \cup C$ and $B \cap C = \O$ implies:


 * $\ds \sum_A z_{p_1} \cdots z_{p_m} = \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}$

Simplify:


 * $\ds \sum_B z_{p_1} \cdots z_{p_m} = z_{n + 1} \sum_D z_{p_1} \cdots z_{p_{m - 1} }$

By definition of elementary symmetric function:

Assemble the preceding equations: