Proving an identity for complete homogenous symmetric polynomials Probably everybody knows the expression:
$$
\sum_{k_1,k_2\ge0}^{k_1+k_2=k}a_1^{k_1}a_2^{k_2}=\frac{a_1^{k+1}-a_2^{k+1}}{a_1-a_2},
$$
where $a_1\ne a_2$ is assumed.
It seems that it can be further generalized to the following statement. Let $a_i$ ($i=1..n)$ be some numbers such that for any $i\ne j$: $a_i\ne a_j$. Then:
$$
\sum_{k_1,k_2,\dots,k_n\ge0}^{\sum_{i=1}^n k_i=k}\prod a_i^{k_i}=\sum_{i=1}^n\frac{a_i^{k+n-1}}{\prod_{j\ne i}(a_i-a_j)}.
$$
Is there a special name for this expansion? What is the simplest way to prove it?
 A: 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.
A: 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.
