[proofplan] We first prove that the products of elementary symmetric functions form a basis by comparing them with the monomial symmetric function basis. The expansion of $e_\lambda$ is triangular: its leading monomial symmetric term is $m_{\lambda'}$, where $\lambda'$ is the conjugate partition, and the coefficient is $1$. Since conjugation is a bijection on partitions of $d$, this gives a unitriangular transition matrix. We then prove the complete homogeneous case by using the formal generating series identity $E(-t)H(t)=1$, which shows that the subalgebra generated by the $h_r$ is the same as the subalgebra generated by the $e_r$ in each degree. [/proofplan] [step:Expand $e_\lambda$ in the monomial symmetric basis] Let $\mathcal{P}(d)$ denote the set of partitions of $d$. For $\mu \in \mathcal{P}(d)$, let $m_\mu$ denote the monomial symmetric function indexed by $\mu$. We use dominance order on $\mathcal{P}(d)$: for partitions $\nu,\rho \in \mathcal{P}(d)$, write $\nu \preceq \rho$ when \begin{align*} \sum_{j=1}^{k}\nu_j \leq \sum_{j=1}^{k}\rho_j \end{align*} for every integer $k \geq 1$, where missing parts are interpreted as $0$; write $\nu \prec \rho$ when $\nu \preceq \rho$ and $\nu \neq \rho$. We use the standard fact built into the definition of $\operatorname{Sym}^d$ that \begin{align*} \{m_\mu : \mu \vdash d\} \end{align*} is a basis of $\operatorname{Sym}^d$. Fix a partition $\lambda = (\lambda_1,\dots,\lambda_\ell)$ of $d$. A monomial appearing in \begin{align*} e_\lambda = e_{\lambda_1}\cdots e_{\lambda_\ell} \end{align*} is obtained by choosing, for each row index $i \in \{1,\dots,\ell\}$, exactly $\lambda_i$ distinct variables from the factor $e_{\lambda_i}$. Thus each such monomial is encoded by a $0$-$1$ matrix $A = (A_{ij})$ with row sums $\lambda_i$. If $\alpha_j := \sum_{i=1}^{\ell} A_{ij}$ denotes the exponent of the variable $x_j$, then the exponent partition $\alpha^+$ is obtained by rearranging the nonzero entries of $\alpha = (\alpha_1,\alpha_2,\dots)$ in weakly decreasing order. We claim that every such exponent partition $\alpha^+$ is dominated by the conjugate partition $\lambda'$. Indeed, for any integer $k \geq 1$, the sum of the $k$ largest column sums of $A$ is bounded above by the largest possible contribution of the rows to any chosen $k$ columns: \begin{align*} \sum_{j=1}^{k} (\alpha^+)_j \leq \sum_{i=1}^{\ell} \min(\lambda_i,k) = \sum_{j=1}^{k} \lambda'_j. \end{align*} The equality on the right is the standard column-counting identity for the Ferrers diagram of $\lambda$. Hence $\alpha^+ \preceq \lambda'$ in dominance order. The exponent partition $\lambda'$ occurs with coefficient $1$. It is realised by the Ferrers matrix whose $i$-th row has ones precisely in columns $1,\dots,\lambda_i$. If a $0$-$1$ matrix with row sums $\lambda_i$ has column sums $\lambda'_j$, then equality holds in the preceding inequality for every $k$, forcing each row of length $\lambda_i$ to place all its ones in the first $\lambda_i$ columns. Thus the Ferrers matrix is the unique contribution to the monomial type $\lambda'$. Therefore \begin{align*} e_\lambda = m_{\lambda'} + \sum_{\mu \prec \lambda'} c_{\lambda\mu} m_\mu \end{align*} for integers $c_{\lambda\mu}$, where $\mu \prec \lambda'$ means that $\mu$ is strictly dominated by $\lambda'$. [guided] The point of this step is to identify the largest monomial symmetric function appearing in $e_\lambda$. Fix $\lambda = (\lambda_1,\dots,\lambda_\ell) \vdash d$. In the product \begin{align*} e_\lambda = e_{\lambda_1}\cdots e_{\lambda_\ell}, \end{align*} the factor $e_{\lambda_i}$ is a sum of squarefree monomials of degree $\lambda_i$. Choosing one monomial from each factor is equivalent to building a $0$-$1$ matrix $A = (A_{ij})$: the entry $A_{ij}$ is $1$ exactly when the variable $x_j$ is selected from the factor $e_{\lambda_i}$. The $i$-th row has exactly $\lambda_i$ ones. Let $\alpha_j := \sum_i A_{ij}$ be the exponent of $x_j$ in the resulting monomial. Let $\alpha^+$ be the partition obtained by sorting the nonzero exponents in weakly decreasing order. We compare $\alpha^+$ with the conjugate partition $\lambda'$. For any $k \geq 1$, the sum of the $k$ largest exponents is the largest total number of ones that can be found in some set of $k$ columns. In row $i$, at most $\min(\lambda_i,k)$ of the $\lambda_i$ ones can lie in those $k$ columns. Hence \begin{align*} \sum_{j=1}^{k}(\alpha^+)_j \leq \sum_{i=1}^{\ell}\min(\lambda_i,k). \end{align*} The right-hand side counts the number of boxes in the first $k$ columns of the Ferrers diagram of $\lambda$, so \begin{align*} \sum_{i=1}^{\ell}\min(\lambda_i,k) = \sum_{j=1}^{k}\lambda'_j. \end{align*} Thus $\alpha^+ \preceq \lambda'$. Now we check the diagonal coefficient. The partition $\lambda'$ is produced by the Ferrers matrix: row $i$ contains ones exactly in columns $1,\dots,\lambda_i$. This gives column sums $\lambda'_1,\lambda'_2,\dots$. Conversely, if a matrix with row sums $\lambda_i$ has column sums exactly $\lambda'_j$, then the preceding inequality must be an equality for every $k$. Equality forces each row to put as many ones as possible into the first $k$ columns for every $k$, and therefore row $i$ must have its ones precisely in columns $1,\dots,\lambda_i$. Hence the Ferrers matrix is the only matrix contributing to $m_{\lambda'}$, so the coefficient of $m_{\lambda'}$ in $e_\lambda$ is $1$. Consequently the expansion has the triangular form \begin{align*} e_\lambda = m_{\lambda'} + \sum_{\mu \prec \lambda'} c_{\lambda\mu}m_\mu, \end{align*} with integer coefficients $c_{\lambda\mu}$. [/guided] [/step] [step:Use triangularity to prove that the elementary products form a basis] Order the partitions of $d$ by any total order refining dominance order. Since the map \begin{align*} \lambda \mapsto \lambda' \end{align*} is a bijection from $\mathcal{P}(d)$ to $\mathcal{P}(d)$, the transition matrix from the ordered family $\{e_\lambda : \lambda \vdash d\}$ to the ordered monomial basis $\{m_\mu : \mu \vdash d\}$ is triangular after reindexing its columns by $\lambda'$. The preceding step shows that every diagonal entry is $1$. Hence this transition matrix is unitriangular and therefore invertible over the coefficient ring. Since the monomial symmetric functions form a basis of $\operatorname{Sym}^d$, the family \begin{align*} \{e_\lambda : \lambda \vdash d\} \end{align*} is also a basis of $\operatorname{Sym}^d$. [/step] [step:Relate elementary and complete homogeneous symmetric functions by generating series] Let $R$ denote the coefficient ring of the symmetric functions, and let $\operatorname{Sym}[[t]]$ denote the ring of formal [power series](/page/Power%20Series) in the indeterminate $t$ with coefficients in the graded $R$-algebra $\operatorname{Sym}$. Define the formal power series $E(t),H(t) \in \operatorname{Sym}[[t]]$ by \begin{align*} E(t) := \sum_{r=0}^{\infty} e_r t^r, \qquad H(t) := \sum_{r=0}^{\infty} h_r t^r, \end{align*} where $e_0 = h_0 = 1$. For an integer $N \geq 1$, let $E_N(t)$ and $H_N(t)$ denote the corresponding finite-variable truncations in the [polynomial ring](/page/Polynomial%20Ring) $R[x_1,\dots,x_N][[t]]$. In the variables $x_1,\dots,x_N$, these series satisfy \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*} Therefore \begin{align*} E_N(-t)H_N(t)=1. \end{align*} Passing coefficientwise to symmetric functions gives \begin{align*} E(-t)H(t)=1. \end{align*} Equating the coefficient of $t^r$ for $r \geq 1$ yields \begin{align*} \sum_{i=0}^{r}(-1)^i e_i h_{r-i}=0. \end{align*} Thus \begin{align*} h_r = \sum_{i=1}^{r}(-1)^{i+1}e_i h_{r-i}. \end{align*} By induction on $r$, each $h_r$ lies in the subalgebra generated by $e_1,\dots,e_r$. The same identity also gives, after solving for $e_r$, \begin{align*} e_r = \sum_{i=1}^{r}(-1)^{i+1}h_i e_{r-i}. \end{align*} By induction on $r$, each $e_r$ lies in the subalgebra generated by $h_1,\dots,h_r$. [/step] [step:Deduce that the complete homogeneous products form a basis] Let $V_d$ denote the span of the family \begin{align*} \{h_\lambda : \lambda \vdash d\} \end{align*} inside $\operatorname{Sym}^d$. The recurrence from the preceding step shows that each $e_r$ is a polynomial in $h_1,\dots,h_r$ whose terms are homogeneous of total degree $r$. Therefore, for every partition $\lambda \vdash d$, the product $e_\lambda$ is a linear combination of products $h_\mu$ with $\mu \vdash d$. Hence \begin{align*} e_\lambda \in V_d \end{align*} for every $\lambda \vdash d$. Since the family $\{e_\lambda : \lambda \vdash d\}$ is a basis of $\operatorname{Sym}^d$, the subspace $V_d$ equals all of $\operatorname{Sym}^d$. Thus the family $\{h_\lambda : \lambda \vdash d\}$ spans $\operatorname{Sym}^d$. It has exactly $|\mathcal{P}(d)|$ elements, which is the rank of the finite free $R$-module $\operatorname{Sym}^d$ because the monomial symmetric functions are indexed by partitions of $d$. The induced $R$-linear endomorphism of $\operatorname{Sym}^d$ that sends the ordered monomial basis to the ordered spanning family $\{h_\lambda : \lambda \vdash d\}$ is surjective; since $\operatorname{Sym}^d$ is finite free, its determinant generates the unit ideal, hence is a unit, so the endomorphism is invertible. Therefore the spanning family \begin{align*} \{h_\lambda : \lambda \vdash d\} \end{align*} is a basis of $\operatorname{Sym}^d$. Combining this with the elementary case proves that both stated families are bases of $\operatorname{Sym}^d$. [/step]