[proofplan] We first prove the inverse identity after restricting to finitely many variables, where the elementary generating polynomial and the complete homogeneous generating series admit complementary product factorizations. In that finite-variable setting, each linear factor cancels with its formal inverse. We then pass to the ring of symmetric functions using the stable specialization system: equality of symmetric functions is detected by all finite-variable specializations, coefficient by coefficient. Finally, expanding the product gives the equivalent coefficient recurrence. [/proofplan]

[step:Prove the inverse identity in finitely many variables]For each integer $n \geq 1$, let \begin{align*} R_n := k[x_1,\dots,x_n]^{S_n} \end{align*} be the ring of symmetric polynomials in $n$ variables over $k$. For $r \geq 0$, let $e_{r,n} \in R_n$ denote the elementary symmetric polynomial of degree $r$, with $e_{0,n}=1$ and $e_{r,n}=0$ for $r>n$. Let $h_{r,n} \in R_n$ denote the complete homogeneous symmetric polynomial of degree $r$, with $h_{0,n}=1$. Define \begin{align*} E_n(t) := \sum_{r=0}^{n} e_{r,n}t^r \in R_n[t], \qquad H_n(t) := \sum_{r=0}^{\infty} h_{r,n}t^r \in R_n[[t]]. \end{align*} By the definitions of elementary and complete homogeneous symmetric polynomials, \begin{align*} E_n(t) = \prod_{j=1}^{n}(1+x_jt), \qquad H_n(t) = \prod_{j=1}^{n}(1-x_jt)^{-1} \end{align*} in $R_n[[t]]$. Substituting $-t$ into $E_n(t)$ gives \begin{align*} E_n(-t)=\prod_{j=1}^{n}(1-x_jt). \end{align*} Therefore, in the commutative formal [power series](/page/Power%20Series) ring $R_n[[t]]$, \begin{align*} H_n(t)E_n(-t) &= \left(\prod_{j=1}^{n}(1-x_jt)^{-1}\right) \left(\prod_{j=1}^{n}(1-x_jt)\right) \\ &=1. \end{align*}[/step]

[guided]We first work with only finitely many variables because the generating functions have a concrete product description there. Fix an integer $n \geq 1$ and define \begin{align*} R_n := k[x_1,\dots,x_n]^{S_n}, \end{align*} the subring of polynomials fixed by the symmetric group $S_n$ acting by permuting the variables. For each $r \geq 0$, the element $e_{r,n} \in R_n$ is the elementary symmetric polynomial of degree $r$, and $h_{r,n} \in R_n$ is the complete homogeneous symmetric polynomial of degree $r$. We use the conventions $e_{0,n}=h_{0,n}=1$ and $e_{r,n}=0$ for $r>n$. The corresponding generating series are \begin{align*} E_n(t) := \sum_{r=0}^{n} e_{r,n}t^r \in R_n[t], \qquad H_n(t) := \sum_{r=0}^{\infty} h_{r,n}t^r \in R_n[[t]]. \end{align*} The elementary generating polynomial factors as \begin{align*} E_n(t)=\prod_{j=1}^{n}(1+x_jt), \end{align*} because choosing the term $x_jt$ from exactly $r$ of the $n$ factors and choosing $1$ from the remaining factors produces precisely all square-free monomials of degree $r$, whose sum is $e_{r,n}$. Similarly, \begin{align*} H_n(t)=\prod_{j=1}^{n}(1-x_jt)^{-1} \end{align*} in $R_n[[t]]$. Indeed, each factor expands as the geometric series \begin{align*} (1-x_jt)^{-1}=\sum_{a=0}^{\infty} x_j^a t^a, \end{align*} which is valid in the formal power series ring because the constant term of $1-x_jt$ is $1$. Multiplying these $n$ geometric series, the coefficient of $t^r$ is the sum of all monomials $x_1^{a_1}\cdots x_n^{a_n}$ with $a_1+\cdots+a_n=r$, exactly $h_{r,n}$. Now substitute $-t$ into $E_n(t)$. This gives \begin{align*} E_n(-t)=\prod_{j=1}^{n}(1-x_jt). \end{align*} Hence \begin{align*} H_n(t)E_n(-t) &= \left(\prod_{j=1}^{n}(1-x_jt)^{-1}\right) \left(\prod_{j=1}^{n}(1-x_jt)\right) \\ &=1. \end{align*} The cancellation is legitimate inside $R_n[[t]]$ because each factor $1-x_jt$ has constant term $1$, hence is a unit in the formal power series ring.[/guided]

[step:Pass the finite-variable identity to the stable symmetric function ring] For $n \geq 2$, define the specialization map \begin{align*} \rho_n: R_n &\to R_{n-1} \\ f(x_1,\dots,x_n) &\mapsto f(x_1,\dots,x_{n-1},0). \end{align*} These maps extend coefficientwise to maps \begin{align*} \rho_n[[t]]: R_n[[t]] \to R_{n-1}[[t]]. \end{align*} The families $(e_{r,n})_{n \geq 1}$ and $(h_{r,n})_{n \geq 1}$ are compatible with these maps, since \begin{align*} \rho_n(e_{r,n})=e_{r,n-1}, \qquad \rho_n(h_{r,n})=h_{r,n-1}. \end{align*} Thus the stable elements $e_r,h_r \in \operatorname{Sym}_k$ are represented by these compatible families, and the series $E(t),H(t) \in \operatorname{Sym}_k[[t]]$ project to $E_n(t),H_n(t)$ for every $n$. We use the equality-detection principle in the inverse-limit model of $\operatorname{Sym}_k$: two elements of $\operatorname{Sym}_k[[t]]$ are equal if, for every power of $t$, their coefficients have the same image in every finite symmetric [polynomial ring](/page/Polynomial%20Ring) $R_n$ under the specialization system. Since $H_n(t)E_n(-t)=1$ in $R_n[[t]]$ for every $n \geq 1$, every finite-variable specialization of every coefficient of $H(t)E(-t)-1$ is zero. Hence the compatible family representing $H(t)E(-t)$ is the constant family representing $1$. Therefore \begin{align*} H(t)E(-t)=1 \end{align*} in $\operatorname{Sym}_k[[t]]$. [/step]

[step:Compare coefficients to obtain the recurrence] Multiplication in $\operatorname{Sym}_k[[t]]$ gives \begin{align*} H(t)E(-t) &= \left(\sum_{a=0}^{\infty} h_a t^a\right) \left(\sum_{i=0}^{\infty} (-1)^i e_i t^i\right) \\ &= \sum_{r=0}^{\infty} \left(\sum_{i=0}^{r} (-1)^i e_i h_{r-i}\right)t^r. \end{align*} The identity $H(t)E(-t)=1$ means that the coefficient of $t^0$ is $1$ and every coefficient of $t^r$ for $r \geq 1$ is $0$. Since $e_0=h_0=1$, the constant coefficient is $1$. Therefore, for every integer $r \geq 1$, \begin{align*} \sum_{i=0}^{r} (-1)^i e_i h_{r-i}=0. \end{align*} This is the equivalent coefficient form of the inverse identity. [/step]