[proofplan] We compare the standard generating series for complete, elementary, and power sum symmetric functions. First, the defining relation between $H(t)$ and $E(t)$, together with the defining property of $\omega$, forces $\omega(H(t))=E(t)$ and hence $\omega(h_n)=e_n$. Then the logarithmic generating series for $H(t)$ and $E(t)$ identify the coefficient-wise action of $\omega$ on the power sums. Finally, multiplicativity of $\omega$ gives the formula for products $p_\lambda$. [/proofplan] [step:Apply $\omega$ to the generating relation for $H(t)$ and $E(t)$] Define the complete and elementary generating series $H(t),E(t) \in \Lambda_{\mathbb{Q}}[[t]]$ by \begin{align*} H(t) := \sum_{n=0}^{\infty} h_n t^n, \qquad E(t) := \sum_{n=0}^{\infty} e_n t^n, \end{align*} with $h_0=e_0=1$. The standard relation between complete and elementary symmetric functions is \begin{align*} H(t)E(-t)=1 \end{align*} in $\Lambda_{\mathbb{Q}}[[t]]$. Extend $\omega$ coefficientwise to a $\mathbb{Q}[[t]]$-algebra automorphism \begin{align*} \omega_t: \Lambda_{\mathbb{Q}}[[t]] &\to \Lambda_{\mathbb{Q}}[[t]], \\ \sum_{n=0}^{\infty} f_n t^n &\mapsto \sum_{n=0}^{\infty} \omega(f_n)t^n. \end{align*} Applying $\omega_t$ to $H(t)E(-t)=1$ gives \begin{align*} \omega_t(H(t))\,\omega_t(E(-t))=1. \end{align*} Since $\omega(e_0)=1$ and $\omega(e_n)=h_n$ for $n\geq 1$, we have \begin{align*} \omega_t(E(-t)) &= \sum_{n=0}^{\infty} \omega(e_n)(-t)^n \\ &= \sum_{n=0}^{\infty} h_n(-t)^n \\ &= H(-t). \end{align*} Therefore \begin{align*} \omega_t(H(t))H(-t)=1. \end{align*} Replacing $t$ by $-t$ in $H(t)E(-t)=1$ gives \begin{align*} H(-t)E(t)=1. \end{align*} The formal [power series](/page/Power%20Series) $H(-t)$ has constant coefficient $1$, so it is invertible in $\Lambda_{\mathbb{Q}}[[t]]$. Hence its inverse is unique, and the two identities above imply \begin{align*} \omega_t(H(t))=E(t). \end{align*} Comparing coefficients of $t^n$ yields \begin{align*} \omega(h_n)=e_n \end{align*} for every $n\geq 1$. [guided] The goal of this step is to turn the defining information $\omega(e_n)=h_n$ into the corresponding formula for $h_n$. We package the two families into formal power series: \begin{align*} H(t) := \sum_{n=0}^{\infty} h_n t^n, \qquad E(t) := \sum_{n=0}^{\infty} e_n t^n, \end{align*} where $h_0=e_0=1$. These series satisfy \begin{align*} H(t)E(-t)=1. \end{align*} Because $\omega$ is an algebra automorphism of $\Lambda_{\mathbb{Q}}$, it extends coefficientwise to formal power series. Define \begin{align*} \omega_t: \Lambda_{\mathbb{Q}}[[t]] &\to \Lambda_{\mathbb{Q}}[[t]], \\ \sum_{n=0}^{\infty} f_n t^n &\mapsto \sum_{n=0}^{\infty} \omega(f_n)t^n. \end{align*} This map fixes the formal variable $t$ and applies $\omega$ only to coefficients. Applying it to the identity $H(t)E(-t)=1$ gives \begin{align*} \omega_t(H(t))\,\omega_t(E(-t))=1. \end{align*} Now compute the second factor. Since $\omega(e_0)=\omega(1)=1=h_0$ and $\omega(e_n)=h_n$ for all $n\geq 1$, we get \begin{align*} \omega_t(E(-t)) &= \omega_t\left(\sum_{n=0}^{\infty} e_n(-t)^n\right) \\ &= \sum_{n=0}^{\infty} \omega(e_n)(-t)^n \\ &= \sum_{n=0}^{\infty} h_n(-t)^n \\ &= H(-t). \end{align*} Thus \begin{align*} \omega_t(H(t))H(-t)=1. \end{align*} On the other hand, replacing $t$ by $-t$ in the original relation $H(t)E(-t)=1$ gives \begin{align*} H(-t)E(t)=1. \end{align*} The series $H(-t)$ has constant coefficient $1$, so it is a unit in the formal power series ring $\Lambda_{\mathbb{Q}}[[t]]$. A unit has a unique inverse. Since both $\omega_t(H(t))$ and $E(t)$ are right inverses of $H(-t)$, they must be equal: \begin{align*} \omega_t(H(t))=E(t). \end{align*} Finally, equality of formal power series is coefficientwise equality, so comparing coefficients of $t^n$ gives \begin{align*} \omega(h_n)=e_n \end{align*} for every $n\geq 1$. [/guided] [/step] [step:Compare logarithmic generating series to compute $\omega(p_n)$] Define the power-sum logarithmic series $P(t) \in t\Lambda_{\mathbb{Q}}[[t]]$ by \begin{align*} P(t) := \sum_{n=1}^{\infty} \frac{p_n t^n}{n}. \end{align*} The standard logarithmic identities are \begin{align*} H(t)=\exp(P(t)), \qquad E(t)=\exp\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{p_n t^n}{n}\right). \end{align*} Because $\omega_t$ is a $\mathbb{Q}$-algebra homomorphism and preserves multiplication, it commutes with the formal exponential series. Hence \begin{align*} \omega_t(H(t)) &= \omega_t(\exp(P(t))) \\ &= \exp(\omega_t(P(t))) \\ &= \exp\left(\sum_{n=1}^{\infty}\frac{\omega(p_n)t^n}{n}\right). \end{align*} From the previous step, $\omega_t(H(t))=E(t)$, so \begin{align*} \exp\left(\sum_{n=1}^{\infty}\frac{\omega(p_n)t^n}{n}\right) = \exp\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{p_n t^n}{n}\right). \end{align*} Both exponents lie in $t\Lambda_{\mathbb{Q}}[[t]]$, and the formal exponential is injective on $t\Lambda_{\mathbb{Q}}[[t]]$ with inverse the formal logarithm. Applying the formal logarithm to both sides gives \begin{align*} \sum_{n=1}^{\infty}\frac{\omega(p_n)t^n}{n} = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{p_n t^n}{n}. \end{align*} Comparing coefficients of $t^n$ and multiplying by $n$ gives \begin{align*} \omega(p_n)=(-1)^{n-1}p_n \end{align*} for every $n\geq 1$. [/step] [step:Use multiplicativity to obtain the formula for $p_\lambda$] Let $\lambda=(\lambda_1,\dots,\lambda_r)$ be a partition. By definition, \begin{align*} p_\lambda := p_{\lambda_1}\cdots p_{\lambda_r}, \qquad |\lambda| := \lambda_1+\cdots+\lambda_r, \qquad \ell(\lambda):=r. \end{align*} Since $\omega$ is an algebra homomorphism, it is multiplicative. Therefore \begin{align*} \omega(p_\lambda) &= \omega(p_{\lambda_1}\cdots p_{\lambda_r}) \\ &= \omega(p_{\lambda_1})\cdots \omega(p_{\lambda_r}) \\ &= \left((-1)^{\lambda_1-1}p_{\lambda_1}\right)\cdots \left((-1)^{\lambda_r-1}p_{\lambda_r}\right) \\ &= (-1)^{(\lambda_1-1)+\cdots+(\lambda_r-1)} p_{\lambda_1}\cdots p_{\lambda_r} \\ &= (-1)^{|\lambda|-\ell(\lambda)}p_\lambda. \end{align*} This is the claimed formula for every partition $\lambda$. [/step]