[proofplan] We classify monomials in $k[x_1,\dots,x_n]$ by the orbit of their exponent vector under the natural action of the symmetric group $S_n$. Each orbit is determined by a unique partition $\lambda$ whose parts are the positive exponents in decreasing order, and the corresponding orbit sum is exactly $m_\lambda^{(n)}$. A homogeneous symmetric polynomial has constant coefficients on every such orbit, so it is a $k$-linear combination of the relevant $m_\lambda^{(n)}$. Distinct partitions give disjoint monomial supports, which gives [linear independence](/page/Linear%20Independence). [/proofplan]

[step:Classify degree $d$ monomials by partitions of $d$ with at most $n$ parts]Let $M_d$ denote the set of exponent vectors of total degree $d$: \begin{align*} M_d := \left\{\alpha = (\alpha_1,\dots,\alpha_n) \in \mathbb{Z}_{\ge 0}^n : \alpha_1 + \cdots + \alpha_n = d\right\}. \end{align*} For $\alpha \in M_d$, define $\lambda(\alpha)$ to be the finite sequence obtained by deleting the zero entries of $\alpha$ and arranging the remaining entries in weakly decreasing order. Then $\lambda(\alpha)$ is a partition of $d$ and $\ell(\lambda(\alpha)) \le n$. Conversely, if $\lambda = (\lambda_1,\dots,\lambda_r)$ is a partition of $d$ with $r \le n$, then the exponent vector \begin{align*} (\lambda_1,\dots,\lambda_r,0,\dots,0) \in \mathbb{Z}_{\ge 0}^n \end{align*} belongs to $M_d$ and has associated partition $\lambda$. Thus partitions of $d$ with length at most $n$ are exactly the possible sorted exponent types of degree $d$ monomials.[/step]

[guided]A monomial in $k[x_1,\dots,x_n]$ is determined by an exponent vector \begin{align*} \alpha = (\alpha_1,\dots,\alpha_n) \in \mathbb{Z}_{\ge 0}^n, \end{align*} and the monomial has total degree $d$ precisely when \begin{align*} \alpha_1 + \cdots + \alpha_n = d. \end{align*} We therefore define \begin{align*} M_d := \left\{\alpha = (\alpha_1,\dots,\alpha_n) \in \mathbb{Z}_{\ge 0}^n : \alpha_1 + \cdots + \alpha_n = d\right\}. \end{align*} The symmetric group permutes variables, so only the multiset of exponents matters for an orbit. To record that multiset in a standard way, take an exponent vector $\alpha \in M_d$, remove its zero entries, and arrange the remaining entries in weakly decreasing order. Call the resulting sequence $\lambda(\alpha)$. Its entries are positive integers, weakly decreasing, and sum to $d$, so $\lambda(\alpha)$ is a partition of $d$. Since $\alpha$ has only $n$ entries, the number of nonzero entries is at most $n$, hence $\ell(\lambda(\alpha)) \le n$. Conversely, suppose $\lambda = (\lambda_1,\dots,\lambda_r)$ is a partition of $d$ with $r \le n$. Padding $\lambda$ by zeros gives the exponent vector \begin{align*} (\lambda_1,\dots,\lambda_r,0,\dots,0) \in \mathbb{Z}_{\ge 0}^n. \end{align*} Its total degree is \begin{align*} \lambda_1 + \cdots + \lambda_r = d, \end{align*} so it lies in $M_d$, and deleting zeros and sorting positive entries recovers $\lambda$. Thus every partition of $d$ with at most $n$ parts occurs as the exponent type of some degree $d$ monomial, and no other exponent types occur.[/guided]

[step:Identify each monomial symmetric polynomial with one exponent orbit] Let $S_n$ act on $M_d$ by permuting coordinates: for $\sigma \in S_n$ and $\alpha \in M_d$, define \begin{align*} \sigma \cdot \alpha := (\alpha_{\sigma^{-1}(1)},\dots,\alpha_{\sigma^{-1}(n)}). \end{align*} For a partition $\lambda$ of $d$ with $\ell(\lambda) \le n$, define \begin{align*} O_\lambda := \{\alpha \in M_d : \lambda(\alpha) = \lambda\}. \end{align*} Then $O_\lambda$ is exactly one $S_n$-orbit in $M_d$: two exponent vectors have the same sorted nonzero entries if and only if one is obtained from the other by permuting coordinates. By the definition of $m_\lambda^{(n)}$, we have \begin{align*} m_\lambda^{(n)} = \sum_{\alpha \in O_\lambda} x_1^{\alpha_1}\cdots x_n^{\alpha_n}. \end{align*} Thus $m_\lambda^{(n)}$ is the orbit sum of the monomials whose exponent type is $\lambda$. [/step]

[step:Write every homogeneous symmetric polynomial as a sum of orbit sums] Let $f \in \operatorname{Sym}_n(k)$ be homogeneous of degree $d$. Since the monomials of total degree $d$ form the standard $k$-basis of the homogeneous degree $d$ part of $k[x_1,\dots,x_n]$, there are unique coefficients $c_\alpha \in k$, indexed by $\alpha \in M_d$, such that \begin{align*} f = \sum_{\alpha \in M_d} c_\alpha x_1^{\alpha_1}\cdots x_n^{\alpha_n}. \end{align*} Because $f$ is symmetric, for every $\sigma \in S_n$ the polynomial obtained by permuting variables by $\sigma$ equals $f$. Comparing coefficients in the standard monomial basis gives \begin{align*} c_{\sigma \cdot \alpha} = c_\alpha \end{align*} for every $\alpha \in M_d$ and every $\sigma \in S_n$. Hence $c_\alpha$ is constant on each orbit $O_\lambda$. For each partition $\lambda$ of $d$ with $\ell(\lambda) \le n$, choose any $\alpha_\lambda \in O_\lambda$ and define $a_\lambda := c_{\alpha_\lambda} \in k$. The constancy on $O_\lambda$ gives \begin{align*} f = \sum_{\lambda \vdash d,\ \ell(\lambda)\le n} a_\lambda \sum_{\alpha \in O_\lambda} x_1^{\alpha_1}\cdots x_n^{\alpha_n} = \sum_{\lambda \vdash d,\ \ell(\lambda)\le n} a_\lambda m_\lambda^{(n)}. \end{align*} Thus the displayed family spans the homogeneous degree $d$ part of $\operatorname{Sym}_n(k)$. [/step]

[step:Use disjoint monomial supports to prove linear independence] Suppose that finitely many coefficients $a_\lambda \in k$, indexed by partitions $\lambda$ of $d$ with $\ell(\lambda) \le n$, satisfy \begin{align*} \sum_{\lambda \vdash d,\ \ell(\lambda)\le n} a_\lambda m_\lambda^{(n)} = 0. \end{align*} The sets $O_\lambda$ are pairwise disjoint because an exponent vector has a unique sorted nonzero exponent sequence. Therefore each monomial \begin{align*} x_1^{\alpha_1}\cdots x_n^{\alpha_n} \end{align*} appears in exactly one of the sums $m_\lambda^{(n)}$. Fix a partition $\mu$ of $d$ with $\ell(\mu) \le n$, and choose $\beta \in O_\mu$. The coefficient of the monomial \begin{align*} x_1^{\beta_1}\cdots x_n^{\beta_n} \end{align*} in the linear combination is exactly $a_\mu$. Since the zero polynomial has coefficient $0$ on every standard monomial, $a_\mu = 0$. As $\mu$ was arbitrary, all coefficients vanish, so the family is linearly independent over $k$. [/step]

[step:Conclude the basis statement for the homogeneous symmetric component] The previous spanning step shows that every homogeneous symmetric polynomial of degree $d$ is a $k$-linear combination of the polynomials \begin{align*} m_\lambda^{(n)} \end{align*} with $\lambda$ a partition of $d$ and $\ell(\lambda) \le n$. The previous independence step shows that this expression is unique. Hence \begin{align*} \{m_\lambda^{(n)} : \lambda \text{ is a partition of } d \text{ and } \ell(\lambda) \le n\} \end{align*} is a $k$-basis of the homogeneous degree $d$ part of $\operatorname{Sym}_n(k)$. [/step]