[proofplan] We prove surjectivity by a leading-monomial elimination algorithm. Given a nonzero symmetric polynomial, its lexicographically largest monomial has weakly decreasing exponents, and a suitable product of elementary symmetric polynomials has exactly that leading monomial with coefficient $1$. Subtracting the corresponding multiple strictly lowers the leading monomial, and the process terminates because only finitely many monomials of bounded total degree can occur. Injectivity follows from the same leading-monomial computation: distinct monomials in the variables $y_1,\dots,y_n$ map to polynomials with distinct leading monomials in $x_1,\dots,x_n$, so the leading term of a nonzero relation cannot cancel. [/proofplan]

[step:Fix the monomial order and compute leading monomials of elementary symmetric products]Order monomials in $k[x_1,\dots,x_n]$ lexicographically with \begin{align*} x_1^{a_1}\cdots x_n^{a_n} >_{\mathrm{lex}} x_1^{b_1}\cdots x_n^{b_n} \end{align*} if, for the first index $i \in \{1,\dots,n\}$ with $a_i \neq b_i$, one has $a_i > b_i$. For a nonzero polynomial $g \in k[x_1,\dots,x_n]$, let $\operatorname{LM}(g)$ denote its largest monomial with nonzero coefficient in this order, and let $\operatorname{LC}(g) \in k$ denote the coefficient of that monomial. For each $r \in \{1,\dots,n\}$, the largest monomial of $e_r^{(n)}$ is \begin{align*} \operatorname{LM}(e_r^{(n)}) = x_1x_2\cdots x_r, \end{align*} and its coefficient is $1_k$. Indeed, among the monomials $x_{i_1}\cdots x_{i_r}$ with $1 \le i_1 < \cdots < i_r \le n$, the lexicographically largest choice is obtained by taking $i_j=j$ for each $j \in \{1,\dots,r\}$. Now let $b_1,\dots,b_n \in \mathbb{N}\cup\{0\}$ and define \begin{align*} E_b := (e_1^{(n)})^{b_1}(e_2^{(n)})^{b_2}\cdots(e_n^{(n)})^{b_n} \in k[x_1,\dots,x_n]. \end{align*} Then \begin{align*} \operatorname{LM}(E_b) = x_1^{b_1+\cdots+b_n}x_2^{b_2+\cdots+b_n}\cdots x_n^{b_n}, \end{align*} and this leading monomial has coefficient $1_k$.[/step]

[guided]The lexicographic order is designed to make the monomial using the earliest variables as large as possible. Thus, in $e_r^{(n)}$, the term $x_1x_2\cdots x_r$ is larger than every other squarefree degree-$r$ monomial: if another term misses some $x_j$ with $j \le r$, then it must use a later variable instead, and the first place where the exponent vector changes makes $x_1x_2\cdots x_r$ larger. We now explain why leading monomials multiply in this special situation. In each $e_r^{(n)}$, the leading term has coefficient $1_k$. Therefore, when multiplying \begin{align*} E_b = (e_1^{(n)})^{b_1}(e_2^{(n)})^{b_2}\cdots(e_n^{(n)})^{b_n}, \end{align*} choosing the leading monomial from every factor gives the monomial \begin{align*} (x_1)^{b_1}(x_1x_2)^{b_2}\cdots(x_1x_2\cdots x_n)^{b_n} = x_1^{b_1+\cdots+b_n}x_2^{b_2+\cdots+b_n}\cdots x_n^{b_n}. \end{align*} Its coefficient is $1_k$, because it is the product of the coefficients $1_k$ from the selected leading terms. No other choice of monomials can produce a lexicographically larger monomial. Replacing the leading monomial in any factor $e_r^{(n)}$ by a smaller monomial decreases the exponent vector at the first variable where the two choices differ, while all earlier variables are unchanged. Since lexicographic order compares from $x_1$ onward, the resulting product is lexicographically smaller than the product obtained by choosing leading monomials everywhere. Hence \begin{align*} \operatorname{LM}(E_b) = x_1^{b_1+\cdots+b_n}x_2^{b_2+\cdots+b_n}\cdots x_n^{b_n}, \end{align*} with leading coefficient $1_k$.[/guided]

[step:Show that the leading monomial of a symmetric polynomial has decreasing exponents] Let $f \in \operatorname{Sym}_n(k)$ be nonzero, and write \begin{align*} \operatorname{LM}(f)=x_1^{a_1}\cdots x_n^{a_n} \end{align*} with $a_i \in \mathbb{N}\cup\{0\}$ for each $i \in \{1,\dots,n\}$. Then \begin{align*} a_1 \ge a_2 \ge \cdots \ge a_n. \end{align*} Indeed, suppose there exists $i \in \{1,\dots,n-1\}$ such that $a_i < a_{i+1}$. Let $\tau_i \in S_n$ be the transposition interchanging $i$ and $i+1$. Since $f$ is symmetric, the coefficient of \begin{align*} x_1^{a_1}\cdots x_i^{a_{i+1}}x_{i+1}^{a_i}\cdots x_n^{a_n} \end{align*} in $f$ is the same as the coefficient of $x_1^{a_1}\cdots x_n^{a_n}$, hence is nonzero. But this transposed monomial is lexicographically larger, because the first changed exponent is at $i$ and $a_{i+1}>a_i$. This contradicts the definition of $\operatorname{LM}(f)$. Therefore no such $i$ exists. [/step]

[step:Eliminate the leading monomial of a symmetric polynomial] Let $f \in \operatorname{Sym}_n(k)$ be nonzero, and write \begin{align*} \operatorname{LM}(f)=x_1^{a_1}\cdots x_n^{a_n}, \qquad \operatorname{LC}(f)=c \in k. \end{align*} By the previous step, $a_1 \ge a_2 \ge \cdots \ge a_n$. Define nonnegative integers \begin{align*} b_r := \begin{cases} a_r-a_{r+1}, & 1 \le r \le n-1,\\ a_n, & r=n. \end{cases} \end{align*} Then \begin{align*} a_i = b_i+b_{i+1}+\cdots+b_n \end{align*} for every $i \in \{1,\dots,n\}$. Hence the polynomial \begin{align*} E_b := (e_1^{(n)})^{b_1}(e_2^{(n)})^{b_2}\cdots(e_n^{(n)})^{b_n} \end{align*} has leading monomial $x_1^{a_1}\cdots x_n^{a_n}$ and leading coefficient $1_k$. Therefore \begin{align*} f_1 := f - cE_b \end{align*} belongs to $\operatorname{Sym}_n(k)$, because both $f$ and $E_b$ are symmetric, and every monomial of $f_1$ is lexicographically smaller than $x_1^{a_1}\cdots x_n^{a_n}$ unless $f_1=0$. [/step]

[step:Iterate the elimination to prove every symmetric polynomial is generated by the elementary symmetric polynomials] We prove that every $f \in \operatorname{Sym}_n(k)$ lies in the subring $k[e_1^{(n)},\dots,e_n^{(n)}] \subseteq k[x_1,\dots,x_n]$. If $f=0$, then $f=0(e_1^{(n)},\dots,e_n^{(n)})$. Suppose $f \neq 0$. Let $D \in \mathbb{N}\cup\{0\}$ be the maximum total degree of a monomial appearing in $f$ with nonzero coefficient. During the elimination procedure from the previous step, when the current leading monomial has exponent vector $(a_1,\dots,a_n)$, the corresponding elementary-symmetric product is homogeneous of total degree \begin{align*} a_1+\cdots+a_n. \end{align*} Thus every monomial introduced by subtracting this product has total degree at most $D$, because the current polynomial never has monomials of total degree exceeding $D$. There are only finitely many monomials in $x_1,\dots,x_n$ of total degree at most $D$. Each nonzero elimination strictly lowers the leading monomial in the lexicographic order. Hence the process terminates after finitely many steps. Summing the elementary-symmetric products subtracted along the way gives a polynomial $F \in k[y_1,\dots,y_n]$ such that \begin{align*} f = F(e_1^{(n)},\dots,e_n^{(n)}). \end{align*} Therefore $\Phi$ is surjective. [/step]

[step:Prove that no nonzero polynomial relation among the elementary symmetric polynomials exists] Let $F \in k[y_1,\dots,y_n]$ be nonzero. Write \begin{align*} F = \sum_{\alpha} c_\alpha y_1^{\alpha_1}\cdots y_n^{\alpha_n}, \end{align*} where the sum is finite, each multi-index $\alpha=(\alpha_1,\dots,\alpha_n)$ lies in $(\mathbb{N}\cup\{0\})^n$, and each displayed coefficient $c_\alpha \in k$ is nonzero. For such an $\alpha$, define \begin{align*} m(\alpha) := x_1^{\alpha_1+\cdots+\alpha_n} x_2^{\alpha_2+\cdots+\alpha_n} \cdots x_n^{\alpha_n}. \end{align*} By the leading-monomial computation in the first step, \begin{align*} \operatorname{LM}\left((e_1^{(n)})^{\alpha_1}\cdots(e_n^{(n)})^{\alpha_n}\right)=m(\alpha), \end{align*} with coefficient $1_k$. The map $\alpha \mapsto m(\alpha)$ is injective. Indeed, if $m(\alpha)=m(\beta)$, then equality of the exponent of $x_n$ gives $\alpha_n=\beta_n$; equality of the exponent of $x_{n-1}$ then gives $\alpha_{n-1}+\alpha_n=\beta_{n-1}+\beta_n$, hence $\alpha_{n-1}=\beta_{n-1}$; continuing upward gives $\alpha_r=\beta_r$ for every $r \in \{1,\dots,n\}$. Choose $\alpha^*$ among the finitely many indices with $c_{\alpha^*}\neq 0$ such that $m(\alpha^*)$ is lexicographically maximal. Since the monomials $m(\alpha)$ are distinct, no other term in \begin{align*} F(e_1^{(n)},\dots,e_n^{(n)}) \end{align*} contributes to the monomial $m(\alpha^*)$ at leading level. Its coefficient is therefore $c_{\alpha^*}\cdot 1_k=c_{\alpha^*}$, which is nonzero. Hence \begin{align*} F(e_1^{(n)},\dots,e_n^{(n)}) \neq 0. \end{align*} Thus $\ker \Phi=\{0\}$, so $\Phi$ is injective. [/step]

[step:Conclude that the substitution homomorphism is an isomorphism] The homomorphism \begin{align*} \Phi: k[y_1,\dots,y_n] &\longrightarrow \operatorname{Sym}_n(k) \\ y_r &\longmapsto e_r^{(n)} \end{align*} is surjective by the existence step and injective by the no-relation step. Therefore $\Phi$ is an isomorphism of $k$-algebras. Equivalently, every symmetric polynomial in $k[x_1,\dots,x_n]$ has a unique expression as a polynomial in $e_1^{(n)},\dots,e_n^{(n)}$. [/step]