[proofplan]
We compare the power-sum generators with the elementary symmetric generators. Newton's identities first show recursively that every $p_m$ lies in the polynomial algebra generated by $e_1,\dots,e_m$, and, because $K$ has characteristic zero, the same identities can be solved recursively to express every $e_m$ in the algebra generated by $p_1,\dots,p_m$. Thus the $p_r$ generate the whole algebra $\operatorname{Sym}_K$. In degree $n$, the products $p_\lambda$ therefore span $\operatorname{Sym}_K^n$, and their number equals the known dimension of that homogeneous component, namely the number of partitions of $n$.
[/proofplan]
[step:Fix the elementary and power-sum notation]
For each integer $m \geq 1$, let $e_m \in \operatorname{Sym}_K^m$ denote the $m$th elementary symmetric function, and set $e_0 := 1 \in \operatorname{Sym}_K^0$. We use the standard grading in which
\begin{align*}
\deg e_m = m,
\qquad
\deg p_m = m.
\end{align*}
For a partition $\lambda = (\lambda_1,\dots,\lambda_\ell)$, define
\begin{align*}
e_\lambda := e_{\lambda_1}\cdots e_{\lambda_\ell}.
\end{align*}
The symmetric-function algebra over $K$ is the polynomial algebra
\begin{align*}
\operatorname{Sym}_K = K[e_1,e_2,\dots],
\end{align*}
graded by $\deg e_m = m$. Hence, for each integer $n \geq 0$, the family
\begin{align*}
\{e_\lambda : \lambda \text{ is a partition of } n\}
\end{align*}
is a $K$-basis of $\operatorname{Sym}_K^n$.
[/step]
[step:Use Newton identities to express each power sum through elementary symmetric functions]
For every integer $m \geq 1$, Newton's identity in $\operatorname{Sym}_K$ is
\begin{align*}
m e_m
=
\sum_{i=1}^{m} (-1)^{i-1} e_{m-i}p_i.
\end{align*}
The term with $i=m$ is $(-1)^{m-1}p_m$, because $e_0=1$. Therefore
\begin{align*}
p_m
=
(-1)^{m-1}m e_m
-
\sum_{i=1}^{m-1}(-1)^{m+i}e_{m-i}p_i.
\end{align*}
We prove by induction on $m$ that $p_m \in K[e_1,\dots,e_m]$. For $m=1$, Newton's identity gives $e_1=p_1$, so $p_1 \in K[e_1]$. Assume that $p_i \in K[e_1,\dots,e_i]$ for every $1 \leq i < m$. Since $e_{m-i} \in K[e_1,\dots,e_m]$ and $p_i \in K[e_1,\dots,e_i] \subset K[e_1,\dots,e_m]$, the displayed formula gives $p_m \in K[e_1,\dots,e_m]$.
[/step]
[step:Solve Newton identities recursively for the elementary generators]
Let $A \subseteq \operatorname{Sym}_K$ be the $K$-subalgebra generated by the power sums:
\begin{align*}
A := K[p_1,p_2,\dots].
\end{align*}
We prove by induction on $m$ that $e_m \in A$ for every integer $m \geq 1$. Since $K$ has characteristic zero, the element $m \cdot 1_K \in K$ is nonzero for every $m \geq 1$, and hence is invertible in the field $K$.
Solving Newton's identity for $e_m$ gives
\begin{align*}
e_m
=
\frac{1}{m}
\sum_{i=1}^{m}(-1)^{i-1}e_{m-i}p_i.
\end{align*}
For $m=1$, this gives $e_1=p_1 \in A$. Assume $e_j \in A$ for every $0 \leq j < m$, where $e_0=1 \in A$. For each $1 \leq i \leq m$, we have $e_{m-i}\in A$ by the induction hypothesis and $p_i \in A$ by definition of $A$. Since $A$ is a $K$-algebra and $1/m \in K$, the displayed formula implies $e_m \in A$.
[guided]
The key point is that Newton's identities are triangular: the identity in degree $m$ contains the new elementary generator $e_m$, the new power sum $p_m$, and only elementary terms $e_j$ with $j<m$ besides $e_m$ itself. Define the $K$-subalgebra generated by the power sums as
\begin{align*}
A := K[p_1,p_2,\dots] \subseteq \operatorname{Sym}_K.
\end{align*}
We want to show that every elementary generator $e_m$ belongs to $A$.
Newton's identity says that, for every $m \geq 1$,
\begin{align*}
m e_m
=
\sum_{i=1}^{m}(-1)^{i-1}e_{m-i}p_i.
\end{align*}
The coefficient $m$ must be invertible to solve this equation for $e_m$. This is exactly where the characteristic-zero hypothesis is used: since $K$ is a field of characteristic zero, the element $m \cdot 1_K$ is nonzero, hence invertible. Therefore
\begin{align*}
e_m
=
\frac{1}{m}
\sum_{i=1}^{m}(-1)^{i-1}e_{m-i}p_i.
\end{align*}
Now proceed by induction on $m$. For $m=1$, the formula gives
\begin{align*}
e_1 = p_1,
\end{align*}
so $e_1 \in A$. Suppose that $e_j \in A$ for every $0 \leq j < m$, with $e_0=1 \in A$. In the formula for $e_m$, each factor $e_{m-i}$ has index $m-i<m$, so it belongs to $A$ by the induction hypothesis, and each $p_i$ belongs to $A$ by definition. Since $A$ is closed under multiplication, addition, and multiplication by scalars from $K$, the whole right-hand side belongs to $A$. Hence $e_m \in A$.
[/guided]
[/step]
[step:Identify the power sums as algebra generators of $\operatorname{Sym}_K$]
From the previous step, every elementary generator $e_m$ belongs to $A=K[p_1,p_2,\dots]$. Since
\begin{align*}
\operatorname{Sym}_K = K[e_1,e_2,\dots],
\end{align*}
we obtain
\begin{align*}
\operatorname{Sym}_K \subseteq A.
\end{align*}
The reverse inclusion $A \subseteq \operatorname{Sym}_K$ holds because each $p_m$ is itself an element of $\operatorname{Sym}_K$. Therefore
\begin{align*}
\operatorname{Sym}_K = K[p_1,p_2,\dots].
\end{align*}
[/step]
[step:Pass from algebra generation to homogeneous spanning in degree $n$]
Fix an integer $n \geq 0$. Since $\operatorname{Sym}_K$ is generated as a $K$-algebra by the homogeneous elements $p_m$ of degree $m$, its degree-$n$ component is spanned by all finite products
\begin{align*}
p_{r_1}\cdots p_{r_\ell}
\end{align*}
with $\ell \geq 0$, $r_j \geq 1$, and
\begin{align*}
r_1+\cdots+r_\ell = n.
\end{align*}
Because multiplication in $\operatorname{Sym}_K$ is commutative, each such product can be reordered into the form $p_\lambda$ for the partition $\lambda$ obtained by arranging $r_1,\dots,r_\ell$ in weakly decreasing order. Hence
\begin{align*}
\{p_\lambda : \lambda \text{ is a partition of } n\}
\end{align*}
spans $\operatorname{Sym}_K^n$ over $K$.
[/step]
[step:Compare the spanning set with the dimension of the homogeneous component]
The family
\begin{align*}
\{e_\lambda : \lambda \text{ is a partition of } n\}
\end{align*}
is a $K$-basis of $\operatorname{Sym}_K^n$. Therefore
\begin{align*}
\dim_K \operatorname{Sym}_K^n
=
\#\{\lambda : \lambda \text{ is a partition of } n\}.
\end{align*}
The spanning family
\begin{align*}
\{p_\lambda : \lambda \text{ is a partition of } n\}
\end{align*}
has exactly the same cardinality. A spanning family in a finite-dimensional [vector space](/page/Vector%20Space) whose cardinality equals the dimension of the space is linearly independent. Hence this family is a $K$-basis of $\operatorname{Sym}_K^n$.
For $n=0$, the only partition is the empty partition $\varnothing$, and $p_{\varnothing}=1$, which is the basis element of $\operatorname{Sym}_K^0=K$. Thus the conclusion holds for every integer $n \geq 0$.
[/step]
