=== Outline === Calculus is used to prove identities (1) and (2) in a single effort.  The tools are Viete's Formulas, the calculus derivative of powers $x^n$ and logarithm $\ln \vert x \vert$, Maclaurin series expansion coefficients, mathematical induction, and Leibniz's Rule: One Variable. === Lemma 1 === {{begin-eqn}} {{eqn | l = \prod_{r \mathop = 1}^n  \paren { 1 + x_r z }       | r = \sum_{m \mathop = 0}^n {\map {e_m} X} z^m }} {{end-eqn}} '''Proof of Lemma 1''' Begin with: {{begin-eqn}} {{eqn | n = 11       | l = \displaystyle \prod_{r \mathop = 1}^n  \paren { x - x_r }        | r = \sum_{i \mathop = 0}^n \paren {-1}^{n-i} {\map {e_{n-i} } X} x^i       | c = Viete's Formulas }} {{end-eqn}} Change variables in (11): $x = -1/z$. Details: Generating Function for Elementary Symmetric Function. {{qed|lemma}} === Lemma 2 === Denote by $D^k \map f z$ the $k$th calculus derivative of $\map f z$. Let: {{begin-eqn}} {{eqn | l = \map G z        | r = \displaystyle \prod_{k \mathop = 1}^n \paren {1 + x_k z}       | c = as in Lemma 1 }} {{eqn | l = \map F z        | r = \dfrac { \map {DQ} z }{\map Q z}       | c = the calculus derivative of $\ln \vert { \map Q z } \vert$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | n = 12       | l = \dfrac { D^m \map G {0} } { m! }        | r = \map { e_m } X }} {{eqn | n = 13       | l = \dfrac { D^m \map F {0} } { m! }        | r = \paren {-1}^m \map {p_{m+1} } X }} {{end-eqn}} '''Proof of Lemma 2''' {{begin-eqn}} {{eqn | l = \map G z       | r = \sum_{m \mathop = 0}^n {\map {e_m} X} z^m       | c = by Lemma 1 }} {{end-eqn}} Then identity (12) holds by Maclaurin series expansion applied to polynomial $G$. Identity (13) will be proved after mathematical induction establishes (14) ''infra''.  Let $\map {\mathbf P} m$ be the statement: {{begin-eqn}} {{eqn | n = 14       | l = D^m \map F z        | r = \sum_{i \mathop = 1}^n \dfrac{ m!\paren {-1}^m  x_i^{m+1} }{ \paren { 1 + x_i z }^{m+1} }       | c = for $m \ge 0$ }} {{end-eqn}} Basis for the induction: $m=0$ By calculus and the definition of $G$: {{begin-eqn}} {{eqn | l = \map F z        | r = \dfrac { D \map Q z}{\map Q z} }} {{eqn | ll = \leadsto        | r = \displaystyle \sum_{i \mathop = 1}^n \dfrac { x_i }{ 1 + x_i z } }} {{end-eqn}} Then $\map {\mathbf P} 0$ is true. Induction step $\map {\mathbf P} m$ implies $\map {\mathbf P} {m+1}$: {{begin-eqn}} {{eqn | l = D^{m+1} \map F z        | r = D \paren {\sum_{i \mathop = 1}^n \dfrac{ m!\paren {-1}^m  x_i^{m+1} }{ \paren { 1 + x_i z }^{m+1} } }       | c = by induction hypothesis $\map {\mathbf P} m$ }} {{eqn | r = \sum_{i \mathop = 1}^n \dfrac{ m! \paren {-1}^m x_i^{m+1} \, \paren {-1} \paren {m+1}  \, x_i}{ \paren { 1 + x_i z }^{m+2} }       | c = by calculus power rule $\dfrac { \d u^{-n} }{ \d z } = \paren { -n } u^{-n-1} \dfrac { \d u }{ \d z }$ }} {{end-eqn}} Simplify to prove $\map {\mathbf P} {m+1}$ is true.  The induction is complete. To prove equation (13), first let $z=0$ in equation (14).  Divide by $m!$ to isolate $\map {p_{m+1} } X$, which proves (13). {{qed|lemma}} === Lemma 3 === {{begin-eqn}} {{eqn | n = 15       | l = \paren {m+1} \map {e_{m+1} } X        | r =  \displaystyle \sum_{r \mathop = 0}^m \paren {-1}^r { \map {e_{m-r} } X } { \map {p_{r+1} } X }       | c = for $m \geq 0$ }} {{end-eqn}} '''Proof of Lemma 3''' Begin with $D \map G z = {\map F z} {\map G z}$ and differentiate $m$ times on variable $z$: {{begin-eqn}} {{eqn | l = D^{m+1} \map G z        | r =  \displaystyle \sum_{r \mathop = 0}^m {\dbinom {m} {r} } D^r {\map F z} \, D^{m-r} {\map G z}       | c = Leibniz's Rule/One Variable }} {{eqn | l = D^{m+1} {\map G 0}        | r = \displaystyle  \sum_{r \mathop = 0}^m  \dbinom {m} {r}  r! \, \paren {-1}^r {\map { p_{r+1} } X}  \,  \paren {m-r}! \,  {\map {e_{m-r} } X}       | c = Evaluate at $ z = 0$ and use equations (12) and (13) in Lemma 2. }} {{eqn | l = \paren {m+1} {\map {e_m} X}        | r = \displaystyle  \sum_{r \mathop = 0}^m  \paren {-1}^r  {\map {e_{m-r} } X} \, {\map {p_{r+1} } X}        | c = Use (12), then collect factorials and simplify. }} {{end-eqn}} {{qed|lemma}} '''Proof of the Theorem''' To prove (1), start with equation (15) in Lemma 3. Change indices via equations $m+1=k$, $r+1=i$.  The summation is from $i=0+1$ to $i=m+1$, which gives range $i=1$ to $k$.  Subscript $m-r$ equals $k-1-i+1$, which simplifies to $k-i$. Then: {{begin-eqn}} {{eqn | n = 16       | l = k \, { \map {e_{k} } X }       | r =  \displaystyle \sum_{i \mathop = 1}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }       | c = for $k \geq 1$, which is equation (1) }} {{end-eqn}} To prove (2), assume $k > n \ge 1$ and $X = \set {x_1,\ldots,x_n}$. Equation (16) implies: {{begin-eqn}} {{eqn | n = 117       | l = 0       | r = k \, { \map {e_{k} } X }       | c = because $ \map {e_{j} } X = 0$ for $j=n+1,\ldots,k$. }} {{eqn | ll = \leadsto       | l = 0       | r = \sum_{i \mathop = 1}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }       | c = by (16) for $k \geq 1$ }} {{eqn | ll = \leadsto       | l = 0       | r = \sum_{i \mathop = k-n}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }       | c = because $ \map {e_{k-i} } X = 0$ when $n+1 \le k-i \le k$ }} {{end-eqn}} Then (2) holds. {{qed}}