On the Wikipedia page for Newton's identities, the following identity is stated
$$
kh_k = \sum_{i=1}^k p_i h_{k-i}, \qquad k \in \mathbb{N} \tag{1}
$$
where $h_k(x_1,x_2, \dots, x_n) = \sum_
{\substack{\ell_1+\dots \ell_n=k\\\ell_i \ge 0}} x_1^{\ell_1}x_2^{\ell_2} \dots x_n^{\ell_n}$ is the Complete homogeneous symmetric polynomial (which is the sum of all monomials of total degree $k$), and where $p_k(x_1,x_2, \dots, x_n)=\sum_{i=1}^{n}x_i^k$ is the Power sum symmetric polynomial. The article doesn't include a proof, so I wanted to attempt one.
The RHS of $(1)$ reminded me of the coefficients obtained from a Cauchy product of two power series, which states
$$
\left(\sum_{i\ge 0} a_i x^i\right) \cdot \left(\sum_{j\ge0} b_j x^j\right) = \sum_{k\ge 0} \left( \sum_{l=0}^k a_l b_{k-l}\right) x^k
$$
which gave me the idea to attempt a solution using generating functions for $h_k, p_k$. For $h_k$, working with formal power series you have the following
$$
\sum_{k\ge 0} h_k(x_1,\ldots,x_n)t^k = \prod_{i=1}^n\sum_{j\ge 0}^\infty(x_it)^j = \prod_{i=1}^n\frac1{1-x_it}
$$
by expanding the product of the sum and using the geometric series formula. For $p_k$ we get
$$
\sum_{k\ge 0}p_k\left(x_1, \dots, x_n\right)t^k =\sum_{i=1}^{n}\sum_{k\ge 0}\left(x_i t\right)^k = \sum_{i=1}^{n} \frac{1}{1-x_i t}
$$
So recalling the derivative of a product formula we get
\begin{align*}
\sum_{k \ge 0} \left[kh_k\left(x_1, \dots, x_n\right)\right]t^k & = t \frac{\mathrm{d}}{\mathrm{d}t} \sum_{k \ge 0} \left[h_k\left(x_1, \dots, x_n\right)\right]t^k\\
& = t \frac{\mathrm{d}}{\mathrm{d}t} \prod_{i=1}^{n} \frac{1}{1-x_i t}\\
& = t\left( \prod_{i=1}^{n} \frac{1}{1-x_i t}\right)\left( \sum_{i=1}^{n} \frac{x_i}{1-x_it}\right)\\
& = \left( \prod_{i=1}^{n} \frac{1}{1-x_i t}\right)\left( \sum_{i=1}^{n} \left[\frac{1}{1-x_it}\color{purple}{-1}\right]\right)\\
& = \left(\sum_{i\ge0}h_i t^i\right)\left(\color{purple}{-n} +\sum_{j\ge0}{p_j} t^j\right)\\
& = \left(\sum_{i\ge0}h_i t^i\right)\left(\sum_{j\ge0}\tilde{p_j} t^j\right)
\end{align*}
where $\tilde{p_j}(x_1, \dots, x_n) = \begin{cases} -n+p_0 =0 & j=0\\ p_j & j\ge 1 \end{cases}$. So by the Cauchy product formula, we have
\begin{align*}
\sum_{k \ge 0} \left[kh_k\right]t^k & = \sum_{k \ge 0} \left[\sum_{\ell=\color{green}{0}}^k \tilde{p_\ell }h_{k-\ell}\right]t^k = \sum_{k \ge 0} \left[ \sum_{\ell=\color{green}{1}}^k p_\ell h_{k-\ell}\right]t^k
\end{align*}
And equating coefficients gives the desired result.
The Wikipedia page mentions that the functions $h_k, p_k$ are studied in commutative algebra and combinatorics, which intuitively makes sense as they carry information about the number of ways you can combine stuff to get a certain quantity. I was wondering if there were some other (possibly simpler) ways of showing this result using some abstract algebra or combinatorics methods. So my question is, does anyone know other ways to show this result? Thanks!
Best Answer
Each $x_1^{\ell_1}x_2^{\ell_2} \dots x_n^{\ell_n}$ monomial in $h_k$ is a monomial in the product $x_j^{s_j}h_{k-s_j}$ where $1\leq s_j\leq\ell_j$ and $1\leq j\leq n$ and there are $\ell_1+\ell_2+\dots\ell_n=k$ such products. Therefore the monomial $x_1^{\ell_1}x_2^{\ell_2} \dots x_n^{\ell_n}$ occurs $k$ times in $\sum_{i=1}^k p_ih_{k-i}$. The result follows.