Complete homogeneous functions expand in power sums with cycle-distribution coefficients

Fix a rank $N\ge 0$, work in $\mathbb{Q}[x_1,\dots ,x_N][\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$, and write $h_d^{(N)}:=h_d(x_1,\dots ,x_N)$ and $p_k^{(N)}:=p_k(x_1,\dots ,x_N)$ for the rank-$N$ specializations. The coefficient of $t^d$ in ${\prod }_{i=1}^N(1-x_{i}t{)}^{-1}$ is ${\sum }_{a_1+\cdots +a_N=d}{x}_1^{a_1}\cdots x_N^{a_N}=h_d^{(N)}$, so ${\sum }_{d\ge 0}{h}_d^{(N)}{t}^d={\prod }_{i=1}^N(1-x_{i}t{)}^{-1}$; taking the formal logarithm gives $\log {\prod }_{i=1}^N(1-x_{i}t{)}^{-1}={\sum }_{i=1}^N{\sum }_{k\ge 1}{x}_i^k{t}^k\mathrm{/}k={\sum }_{k\ge 1}{p}_k^{(N)}{t}^k\mathrm{/}k$, and applying the formal exponential (with $\exp (\log U)=U$ for $U\in 1+t\mathbb{Q}[x_1,\dots ,x_N][\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$ and $\exp (A+B)=\exp (A)\exp (B)$ for series $A,B$ with zero constant term) yields ${\prod }_{i=1}^N(1-x_{i}t{)}^{-1}=\exp ({\sum }_{k\ge 1}{p}_k^{(N)}{t}^k\mathrm{/}k)$. Expanding, $\exp \left({\sum }_{k\ge 1}{p}_k^{(N)}{t}^k\mathrm{/}k\right)={\prod }_{k\ge 1}\exp \left(p_k^{(N)}{t}^k\mathrm{/}k\right)$, and in degree $t^d$ only the factors with $k\le d$ contribute, so $h_d^{(N)}$ equals the sum of ${\prod }_{k\ge 1}(p_k^{(N)}{)}^{m_k}/(k^{m_k}{m}_k!)$ over all tuples $(m_k{)}_{k\ge 1}$ of nonnegative integers with ${\sum }_{k}k{m}_k=d$; grouping the tuple by the partition $\rho$ with $m_k(\rho )=m_k$, and using ${\prod }_k{p}_k^{m_k}=p_{\rho }$ and ${\prod }_k{k}^{m_k}{m}_k!=z_{\rho }$, gives $h_d^{(N)}={\sum }_{\rho ⊢d}{p}_{\rho }^{(N)}\mathrm{/}{z}_{\rho }$.

For $M\ge N$ the transition map that sets $x_{N+1},\dots ,x_M$ to zero sends $h_d^{(M)}\mapsto h_d^{(N)}$ and $p_k^{(M)}\mapsto p_k^{(N)}$, by their finite-rank definitions; hence the rank-$N$ identities of step 1.1 are the projections of a single compatible sequence of degree-$d$ symmetric polynomials. Therefore $h_d={\sum }_{\rho ⊢d}{p}_{\rho }\mathrm{/}{z}_{\rho }$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}^d$ for every $d\ge 0$, since two compatible sequences with equal projections are equal by [F1].

Let $\lambda =({\lambda }_1,\dots ,{\lambda }_k)⊢n$. By definition $h_{\lambda }={\prod }_{j=1}^k{h}_{{\lambda }_j}$ [F2], and applying step 2.1 to each part gives $h_{{\lambda }_j}={\sum }_{{\rho }^{(j)}⊢{\lambda }_j}{p}_{{\rho }^{(j)}}\mathrm{/}{z}_{{\rho }^{(j)}}$; multiplying these $k$ finite sums, $h_{\lambda }={\sum }_{({\rho }^{(1)},\dots ,{\rho }^{(k)})}{\prod }_{j=1}^k{p}_{{\rho }^{(j)}}\mathrm{/}{z}_{{\rho }^{(j)}}$, the sum over all $k$-tuples of partitions with ${\rho }^{(j)}⊢{\lambda }_j$.

The monomial ${\prod }_j{p}_{{\rho }^{(j)}}$ equals $p_{\rho }$ exactly when merging the parts of ${\rho }^{(1)},\dots ,{\rho }^{(k)}$ gives the multiset of parts of $\rho$, that is, when ${\sum }_j{m}_i^{(j)}=m_i(\rho )$ for every $i$, where $m_i^{(j)}:=m_i({\rho }^{(j)})$; a $k$-tuple is uniquely recovered from its matrix $(m_i^{(j)})$ by listing $m_i^{(j)}$ copies of each $i$ in decreasing order, and the further condition ${\sum }_{i}i{m}_i^{(j)}={\lambda }_j$ records that ${\rho }^{(j)}⊢{\lambda }_j$. Hence the coefficient of $p_{\rho }$ in $h_{\lambda }$ equals ${\sum }_{(m_i^{(j)})}{\prod }_{j}1\mathrm{/}{z}_{{\rho }^{(j)}}$, the sum over all matrices with those two properties; and for such a matrix ${\sum }_j{m}_i^{(j)}=m_i(\rho )$ gives ${\prod }_j{i}^{m_i^{(j)}}=i^{m_i(\rho )}$, so ${\prod }_{j}1\mathrm{/}{z}_{{\rho }^{(j)}}=({\prod }_i{m}_i(\rho )!\mathrm{/}{\prod }_{i,j}{m}_i^{(j)}!)/z_{\rho }$. Therefore the coefficient of $p_{\rho }$ in $h_{\lambda }$ is $N(\lambda ,\rho )\mathrm{/}{z}_{\rho }$, because the sum displayed in the statement counts, for each $i$ independently, the assignments of the $m_i(\rho )$ distinct cycles of length $i$ of a fixed permutation of cycle type $\rho$ to the $k$ labelled rows with the multiplicities $m_i^{(j)}$. Since $\{p_{\rho }:\rho ⊢n\}$ is a $\mathbb{Q}$-basis of ${\mathrm{\Lambda }}_{\mathbb{Q}}^n$ [F4], the coefficient comparison gives $h_{\lambda }={\sum }_{\rho ⊢n}N(\lambda ,\rho )p_{\rho }\mathrm{/}{z}_{\rho }$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}^n$.

For $n=0$ one has $\lambda =\rho =\mathrm{\varnothing }$, $k=0$, and the empty product conventions give $h_{\mathrm{\varnothing }}=1=N(\mathrm{\varnothing },\mathrm{\varnothing })p_{\mathrm{\varnothing }}\mathrm{/}{z}_{\mathrm{\varnothing }}$, while for $\lambda =(d)$ the single row must receive every cycle, so $N((d),\rho )=1$ for every $\rho ⊢d$ and step 4.1 gives $h_d={\sum }_{\rho ⊢d}{p}_{\rho }\mathrm{/}{z}_{\rho }$; combined with step 2.1 this is the stated one-row case for every $d\ge 0$, including $d=0$, where both sides equal $1$. The general identity of the statement now follows from step 4.1 in all cases, with the empty partition handled by the computation just given.