[proofplan] We first prove both identities in the [polynomial ring](/page/Polynomial%20Ring) of symmetric polynomials in finitely many variables. The proof uses the generating functions for elementary and complete homogeneous symmetric polynomials and compares coefficients after taking formal logarithmic derivatives. Since each identity is homogeneous of degree $n$, passing to the stable limit of symmetric polynomials gives the corresponding identity in $\operatorname{Sym}$. [/proofplan] [step:Prove the elementary identity in finitely many variables] Fix an integer $m \geq 1$, and let \begin{align*} R_m := \mathbb{Z}[x_1,\dots,x_m]. \end{align*} For $r \geq 0$, let $e_{r,m} \in R_m$ be the $r$-th [elementary symmetric polynomial](/page/Elementary%20Symmetric%20Polynomial) in $x_1,\dots,x_m$, with $e_{0,m}=1$ and $e_{r,m}=0$ for $r>m$. For $r \geq 1$, let \begin{align*} p_{r,m} := \sum_{i=1}^{m} x_i^r \in R_m \end{align*} be the $r$-th [power-sum symmetric polynomial](/page/Power%20Sum%20Symmetric%20Polynomial). Define the elementary generating series $E_m(t) \in R_m[[t]]$ by \begin{align*} E_m(t) := \prod_{i=1}^{m}(1+x_i t)=\sum_{r=0}^{m} e_{r,m}t^r. \end{align*} Since $E_m(t)$ has constant term $1$, it is invertible in $R_m[[t]]$. Differentiating the finite product in the formal [power series](/page/Power%20Series) ring $R_m[[t]]$ gives \begin{align*} E_m'(t) &=E_m(t)\sum_{i=1}^{m}\frac{x_i}{1+x_i t}. \end{align*} For each $i$, the inverse of $1+x_i t$ in $R_m[[t]]$ is the geometric series $\sum_{a=0}^{\infty}(-1)^a x_i^a t^a$. Therefore \begin{align*} \sum_{i=1}^{m}\frac{x_i}{1+x_i t} &=\sum_{i=1}^{m}\sum_{a=0}^{\infty}(-1)^a x_i^{a+1}t^a \\ &=\sum_{q=1}^{\infty}(-1)^{q-1}p_{q,m}t^{q-1}. \end{align*} Hence \begin{align*} E_m'(t) &=E_m(t)\sum_{q=1}^{\infty}(-1)^{q-1}p_{q,m}t^{q-1}. \end{align*} Using $E_m(t)=\sum_{r=0}^{m} e_{r,m}t^r$, the coefficient of $t^{n-1}$ on the right-hand side is \begin{align*} \sum_{r=0}^{n-1}(-1)^{n-r-1}e_{r,m}p_{n-r,m}. \end{align*} The coefficient of $t^{n-1}$ in $E_m'(t)$ is $n e_{n,m}$. Thus \begin{align*} n e_{n,m} &=\sum_{r=0}^{n-1}(-1)^{n-r-1}e_{r,m}p_{n-r,m}. \end{align*} Multiplying by $(-1)^n$ gives \begin{align*} \sum_{r=0}^{n-1}(-1)^r e_{r,m}p_{n-r,m}+(-1)^n n e_{n,m}=0. \end{align*} [guided] Fix an integer $m \geq 1$ and work in the polynomial ring \begin{align*} R_m := \mathbb{Z}[x_1,\dots,x_m]. \end{align*} The point of working with finitely many variables is that the elementary generating function is an honest finite product. For $r \geq 0$, let $e_{r,m}$ be the $r$-th [elementary symmetric polynomial](/page/Elementary%20Symmetric%20Polynomial) in $x_1,\dots,x_m$, with $e_{0,m}=1$ and $e_{r,m}=0$ for $r>m$. For $r \geq 1$, define the [power-sum symmetric polynomial](/page/Power%20Sum%20Symmetric%20Polynomial) \begin{align*} p_{r,m} := \sum_{i=1}^{m} x_i^r. \end{align*} Now define the formal power series \begin{align*} E_m(t) := \prod_{i=1}^{m}(1+x_i t)=\sum_{r=0}^{m} e_{r,m}t^r \end{align*} in $R_m[[t]]$. Its constant term is $1$, so it has a multiplicative inverse in $R_m[[t]]$. This justifies the logarithmic derivative computation entirely inside formal power series. Differentiate the finite product using the product rule: \begin{align*} E_m'(t) &=\sum_{j=1}^{m}x_j\prod_{\substack{1\leq i\leq m\\ i\neq j}}(1+x_i t). \end{align*} Factoring out $E_m(t)=\prod_{i=1}^{m}(1+x_i t)$ gives \begin{align*} E_m'(t) &=E_m(t)\sum_{j=1}^{m}\frac{x_j}{1+x_j t}. \end{align*} This is valid because each $1+x_j t$ has constant term $1$ and is therefore invertible in $R_m[[t]]$. Next expand each inverse as a formal geometric series: \begin{align*} \frac{1}{1+x_j t} &=\sum_{a=0}^{\infty}(-1)^a x_j^a t^a. \end{align*} Multiplying by $x_j$ and summing over $j$ gives \begin{align*} \sum_{j=1}^{m}\frac{x_j}{1+x_j t} &=\sum_{j=1}^{m}\sum_{a=0}^{\infty}(-1)^a x_j^{a+1}t^a \\ &=\sum_{q=1}^{\infty}(-1)^{q-1}\left(\sum_{j=1}^{m}x_j^q\right)t^{q-1} \\ &=\sum_{q=1}^{\infty}(-1)^{q-1}p_{q,m}t^{q-1}. \end{align*} Thus \begin{align*} E_m'(t) &=\left(\sum_{r=0}^{m}e_{r,m}t^r\right) \left(\sum_{q=1}^{\infty}(-1)^{q-1}p_{q,m}t^{q-1}\right). \end{align*} We now compare the coefficient of $t^{n-1}$. On the left, \begin{align*} E_m'(t) &=\sum_{r=1}^{m} r e_{r,m}t^{r-1}, \end{align*} so the coefficient of $t^{n-1}$ is $n e_{n,m}$, with the convention that $e_{n,m}=0$ if $n>m$. On the right, a term $e_{r,m}t^r$ can pair with $(-1)^{q-1}p_{q,m}t^{q-1}$ only when $r+q-1=n-1$, or equivalently $q=n-r$. Since $q\geq 1$, this forces $0\leq r\leq n-1$. Hence the coefficient is \begin{align*} \sum_{r=0}^{n-1}(-1)^{n-r-1}e_{r,m}p_{n-r,m}. \end{align*} Equating coefficients gives \begin{align*} n e_{n,m} &=\sum_{r=0}^{n-1}(-1)^{n-r-1}e_{r,m}p_{n-r,m}. \end{align*} Multiplying both sides by $(-1)^n$ yields \begin{align*} \sum_{r=0}^{n-1}(-1)^r e_{r,m}p_{n-r,m}+(-1)^n n e_{n,m}=0. \end{align*} This is the first Newton identity in $m$ variables. [/guided] [/step] [step:Prove the complete homogeneous identity in finitely many variables] For $r \geq 0$, let $h_{r,m}\in R_m$ be the $r$-th [complete homogeneous symmetric polynomial](/page/Complete%20Homogeneous%20Symmetric%20Polynomial) in $x_1,\dots,x_m$, with $h_{0,m}=1$. Define the complete homogeneous generating series $H_m(t)\in R_m[[t]]$ by \begin{align*} H_m(t):=\prod_{i=1}^{m}(1-x_i t)^{-1}=\sum_{r=0}^{\infty}h_{r,m}t^r. \end{align*} Since each $1-x_i t$ has constant term $1$, this product is well-defined in $R_m[[t]]$. Taking the formal logarithmic derivative gives \begin{align*} H_m'(t) &=H_m(t)\sum_{i=1}^{m}\frac{x_i}{1-x_i t}. \end{align*} Using the geometric expansion $(1-x_i t)^{-1}=\sum_{a=0}^{\infty}x_i^a t^a$, we obtain \begin{align*} \sum_{i=1}^{m}\frac{x_i}{1-x_i t} &=\sum_{i=1}^{m}\sum_{a=0}^{\infty}x_i^{a+1}t^a \\ &=\sum_{q=1}^{\infty}p_{q,m}t^{q-1}. \end{align*} Therefore \begin{align*} H_m'(t) &=H_m(t)\sum_{q=1}^{\infty}p_{q,m}t^{q-1}. \end{align*} Comparing coefficients of $t^{n-1}$ gives \begin{align*} n h_{n,m} &=\sum_{r=0}^{n-1}h_{r,m}p_{n-r,m}. \end{align*} Renaming the index by $q=n-r$ gives \begin{align*} \sum_{q=1}^{n}p_{q,m}h_{n-q,m} &=n h_{n,m}. \end{align*} [guided] Fix the same integer $m \geq 1$ and continue working in \begin{align*} R_m := \mathbb{Z}[x_1,\dots,x_m]. \end{align*} For $r \geq 0$, let $h_{r,m}$ denote the $r$-th [complete homogeneous symmetric polynomial](/page/Complete%20Homogeneous%20Symmetric%20Polynomial) in $x_1,\dots,x_m$, with $h_{0,m}=1$. The complete homogeneous generating series is useful because allowing repetitions of variables turns into a product of geometric series. Define \begin{align*} H_m(t):=\prod_{i=1}^{m}(1-x_i t)^{-1}=\sum_{r=0}^{\infty}h_{r,m}t^r \end{align*} as an element of $R_m[[t]]$. Each factor $1-x_i t$ has constant term $1$, so each inverse $(1-x_i t)^{-1}$ exists in the formal power series ring, and the finite product is well-defined. We take the formal logarithmic derivative. Since $H_m(t)$ is a finite product of invertible formal power series, the product rule gives \begin{align*} H_m'(t) &=H_m(t)\sum_{i=1}^{m}\frac{x_i}{1-x_i t}. \end{align*} The sign is positive because differentiating $(1-x_i t)^{-1}$ gives $x_i(1-x_i t)^{-2}$, and after factoring out $(1-x_i t)^{-1}$ from the whole product, the remaining contribution is $x_i/(1-x_i t)$. Now expand each inverse by the formal geometric series \begin{align*} (1-x_i t)^{-1} &=\sum_{a=0}^{\infty}x_i^a t^a. \end{align*} Multiplying by $x_i$ and summing over the variable index $i$ gives \begin{align*} \sum_{i=1}^{m}\frac{x_i}{1-x_i t} &=\sum_{i=1}^{m}\sum_{a=0}^{\infty}x_i^{a+1}t^a \\ &=\sum_{q=1}^{\infty}\left(\sum_{i=1}^{m}x_i^q\right)t^{q-1} \\ &=\sum_{q=1}^{\infty}p_{q,m}t^{q-1}, \end{align*} where $p_{q,m}=\sum_{i=1}^{m}x_i^q$ is the $q$-th [power-sum symmetric polynomial](/page/Power%20Sum%20Symmetric%20Polynomial). Hence \begin{align*} H_m'(t) &=\left(\sum_{r=0}^{\infty}h_{r,m}t^r\right) \left(\sum_{q=1}^{\infty}p_{q,m}t^{q-1}\right). \end{align*} Compare coefficients of $t^{n-1}$. On the left, \begin{align*} H_m'(t) &=\sum_{r=1}^{\infty}r h_{r,m}t^{r-1}, \end{align*} so the coefficient of $t^{n-1}$ is $n h_{n,m}$. On the right, the product term $h_{r,m}t^r$ pairs with $p_{q,m}t^{q-1}$ exactly when $r+q-1=n-1$, equivalently $q=n-r$. Since $q\geq 1$, this means $0\leq r\leq n-1$. Therefore the coefficient on the right is \begin{align*} \sum_{r=0}^{n-1}h_{r,m}p_{n-r,m}. \end{align*} Equating coefficients in $R_m[[t]]$ gives \begin{align*} n h_{n,m} &=\sum_{r=0}^{n-1}h_{r,m}p_{n-r,m}. \end{align*} Finally set $q=n-r$. As $r$ runs from $0$ to $n-1$, the index $q$ runs from $n$ down to $1$, so reordering the finite sum gives \begin{align*} \sum_{q=1}^{n}p_{q,m}h_{n-q,m} &=n h_{n,m}. \end{align*} This is the complete homogeneous Newton identity in $m$ variables. [/guided] [/step] [step:Pass the finite-variable identities to the ring of symmetric functions] For each $m\geq 1$, let $S_m$ denote the symmetric group on the set $\{1,\dots,m\}$, acting on $R_m$ by permuting the variables $x_1,\dots,x_m$. Let $R_m^{S_m}$ denote the subring of $S_m$-invariant polynomials. Define the specialization homomorphism $\rho_m:\operatorname{Sym}\to R_m^{S_m}$ as the map sending a symmetric function to its symmetric polynomial in the variables $x_1,\dots,x_m$. By definition of the standard bases of $\operatorname{Sym}$, \begin{align*} \rho_m(e_r)&=e_{r,m},& \rho_m(h_r)&=h_{r,m},& \rho_m(p_r)&=p_{r,m}. \end{align*} Fix $n\geq 1$. Apply $\rho_m$ to the two desired identities. The finite-variable identities proved above show that the images under $\rho_m$ are zero for every $m\geq 1$. Each side is homogeneous of degree $n$ in $\operatorname{Sym}$. Let $\lambda=(\lambda_1,\dots,\lambda_k)$ denote a partition of $n$, and let $\ell(\lambda):=k$ denote the number of parts of $\lambda$. Let $m_\lambda\in\operatorname{Sym}$ denote the [monomial symmetric function](/page/Monomial%20Symmetric%20Function) indexed by $\lambda$. The degree-$n$ component of $\operatorname{Sym}$ has the monomial symmetric basis indexed by partitions $\lambda$ of $n$. Every such partition satisfies $\ell(\lambda)\leq n$, so for any $m\geq n$ the specialization $\rho_m$ sends each degree-$n$ monomial symmetric function $m_\lambda$ to the corresponding nonzero monomial symmetric polynomial in $m$ variables, and these specialized monomial symmetric polynomials remain linearly independent. Hence $\rho_m$ is injective on the degree-$n$ component for every $m\geq n$. Therefore the two homogeneous degree-$n$ differences themselves are zero in $\operatorname{Sym}$. Hence, for every $n\geq 1$, \begin{align*} \sum_{r=0}^{n-1}(-1)^r e_rp_{n-r}+(-1)^n n e_n&=0,\\ \sum_{r=1}^{n}p_rh_{n-r}&=n h_n. \end{align*} This proves both Newton identities in the ring of symmetric functions. [/step]