It is sufficient to consider the case $a_n = 1$: :$\displaystyle \map P x = \prod_{k \mathop = 1}^n \paren {x - z_k}$ The proof proceeds by induction. Let $\map {\Bbb P} n$ be the statement that the identity below holds for all sets $\set {z_1, \ldots, z_n}$. {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^n \paren {x - z_j}        | r = x^n + \sum_{j \mathop = 1}^n \paren {-1}^{n - j + 1} e_{n - j + 1} \paren {\set {z_1, \ldots, z_n} } \, x^{j - 1} }} {{eqn | r = x^n + \paren {-1} \, e_1 \paren {\set {z_1, \ldots, z_n} } \, x^{n - 1} + \paren {-1}^2 \, e_2 \paren {\set {z_1, \ldots, z_n} } \, x^{n - 2} + \cdots + \paren {-1}^n e_n \paren {\set {z_1, \ldots, z_n} } }} {{end-eqn}} Basis for the Induction: $\map {\Bbb P} 1$ holds because $\map {e_1} {\set {z_1} } = z_1$. Induction Step $\map {\Bbb P} n$ implies $\map {\Bbb P} {n + 1}$: Assume $\map {\Bbb P} n$ holds and $n \ge 1$. Let for given values $\set {z_1, \ldots, z_n, z_{n + 1} }$: :$\displaystyle \map Q x = \paren {x - z_{n + 1} } \prod_{k \mathop = 1}^n \paren {x - z_k}$ Expand the right side product above using induction hypothesis $\map {\Bbb P} n$. Then $\map Q x$ equals $x^{n + 1}$ plus terms for $x^{j - 1}$, $1 \le j \le n + 1$. If $j = 1$, then one term occurs for $x^{j - 1}$: :$\displaystyle \paren {-x_{n + 1} } \, \paren {\paren {-1}^{n - 1 + 1} \map {e_{n - 1 + 1} } {\set {z_1, \ldots, z_n} } x^{1 - 1} } = \paren {-1}^{n + 1} \map {e_n} {\set {z_1, \ldots, z_n, z_{n + 1} } }$ If $2 \le j \le n + 1$, then two terms $T_1$ and $T_2$ occur for $x^{j - 1}$: {{begin-eqn}} {{eqn | l = T_1        | r = \paren x \paren {\paren {-1}^{n - j + 2} \, \map {e_{n - j + 2} } {\set {z_1, \ldots, z_n} } x^{j - 2} } }} {{eqn | l = T_2        | r = \paren {-z_{n + 1 } } \, \paren {\paren {-1}^{n - j + 1} \, \map {e_{n - j + 1} } {\set {z_1, \ldots, z_n} } x^{j - 1} } }} {{end-eqn}} The coefficient $c$ of $x^{j - 1}$ for $2 \le j \le n + 1$ is: {{begin-eqn}} {{eqn | l = c       | r = \dfrac {T_1 + T_2} {x^{j - 1} } }} {{eqn | r = \paren {-1}^m \, \map {e_m} {\set {z_1, \ldots, z_n} } + \paren {-1}^m \, \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } z_{n + 1}        | c = where $m = n - j + 2$. }} {{end-eqn}} Use recursion identity to simplify the expression for $c$: {{begin-eqn}} {{eqn | l = \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }       | r = z_{n + 1} \, \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} } }} {{eqn | ll= \leadsto       | l = c        | r = \paren {-1}^{n - j + 2} \, \map {e_{n - j + 2} } {\set {z_1, \ldots, z_n, z_{n + 1} } } }} {{end-eqn}} Thus $\map {\Bbb P} {n + 1}$ holds and the induction is complete.  Set equal the two identities for $\map P x$: :$\displaystyle x^n + \sum_{k \mathop = 0}^{n - 1} a_k x^k = x^n + \paren {-1} \, \map {e_1} {\set {z_1, \ldots, z_n} } x^{n - 1} + \paren {-1}^2 \, \map {e_2} {\set {z_1, \ldots, z_n} } x^{n - 2} + \cdots + \paren {-1}^n \map {e_n} {\set {z_1, \ldots, z_n} }$ Linear independence of the powers $1, x, x^2, \ldots$ implies polynomial coefficients match left and right. Hence the coefficient $a_k$ of $x^k$ on the {{LHS}} matches $\paren {-1}^{n - k} \, \map {e_{n - k} } {\set {z_1, \ldots, z_n} }$ on the {{RHS}}. {{qed}}  {{proofread}}