Proving an identity for complete homogenous symmetric polynomials

I don't know if there is a name for this expansion, but this is the complete homogenous symmetric polynomial.

Edit: Here's a proof that uses partial fraction decomposition.

First note $$h_k = \sum_{k_1,k_2,\dots,k_n\ge0}^{\sum_{i=1}^n k_i=k}\prod_i a_i^{k_i}$$ has the generating function $$\sum_{k=0}^\infty h_k t^k = \prod_{i=1}^n \frac{1}{1 - a_it}.$$ So we need to find a way to extract the coefficient of $t^k$ from the right-hand side.

Assuming none of the $a_i$'s are equal, this is a rational function in $t$ with $n$ distinct roots, so we can apply a partial fraction decomposition:

$$\prod_{i=1}^n \frac{1}{1 - a_it} = \sum_{i=1}^n \frac{c_i}{1 - a_it}$$

for some $c_i$ to be determined. Multiplying through by the denominator of the left gives

$$1 = \sum_{i=1}^n c_i\prod_{j \neq i} (1 - a_jt).$$

To find the $c_i$, set $t = 1/a_i$. Then each term vanishes except for the one with $c_i$. This gives

$$1 = c_i\prod_{j \neq i} (1 - a_j/a_i)$$

$$ c_i = \frac{1}{\prod_{j \neq i} (1 - a_j/a_i)} = \frac{a_i^{n-1}}{\prod_{j \neq i} (a_i - a_j)}$$

Then $$\sum_{k=0}^\infty h_k t^k = \sum_{i=1}^n \frac{c_i}{1 - a_it} = \sum_{i=1}^n \sum_{k=0}^\infty c_i a_i^k t^k$$

and so the coefficient of $t^k$ is

$$h_k = \sum_{i=1}^n c_i a_i^k = \sum_{i=1}^n \frac{a_i^{n+k-1}}{\prod_{j \neq i} (a_i - a_j)}$$

Nice work finding this pretty identity, btw! I am sure it is 'well-known' but I had not seen it.

I don't think it has a special name, but the implicated concepts do.

Let $f_t(a)=\frac{1}{1-ta}$. Then the $(n-1)$th divided difference is $$\Delta^{n-1}f_t[a_1,a_2,\ldots,a_n]=\frac{t^{n-1}}{\prod_{i=1}^n(1-a_it)}$$ as may be checked by induction on $n$; the right hand side is $t^{n-1}$ times the generating function of the complete homogeneous symmetric polynomials. But we also have $$\Delta^{n-1}f_t[a_1,a_2,\ldots,a_n]=\sum_{i=1}^n\frac{f_t(a_i)}{\prod_{j\neq i}(a_j-a_i)}=\sum_{i=1}^n\frac{(1-a_it)^{-1}}{\prod_{j\neq i}(a_j-a_i)}$$ (this is a standard identity in the theory of divided differences), so $$\frac{t^{n-1}}{\prod_{i=1}^n(1-a_it)}=\sum_{i=1}^n\frac{(1-a_it)^{-1}}{\prod_{j\neq i}(a_j-a_i)}\text{.}$$ Taking the coefficient of $t^{k+n-1}$ on both sides gives the desired result.