Generalization of $\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$ I thought about a generalization for the formula
$$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$$
It can be written as
$$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1} = \sum_{i + j = n - 1}x^iy^j$$
So we would like to generalize:
$$\sum_{i_1 + i_2 + i_3 + ... +i_k = n - 1}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}$$
For example"
$$\sum_{i + j + k = n - 1}{x}^{i}{y}^{j}{z}^{k} = \sum_{k=0}^{n - 1}{z}^{k} 
\sum_{i + j  = n - k - 1}{x}^{i}{y}^{j} = \sum_{k=0}^{n - 1}{z}^{k} 
\frac{x^{n - k} - y^{n - k}}{x -y} = \frac{1}{x-y} \left(\frac{x^{n + 1} - z^{n + 1}}{x - z} - \frac{y^{n + 1} - z^{n + 1}}{y - z}\right)$$
It seems that the generalized expression is the divided difference of $x^{n + k - 2}$ in the points $x_1, x_2, \ldots, x_k$.
Does anyone have an idea how to prove it?

I think that this formula is what you are looking for. If $\mathbf x = (x_1,\dotsc,x_r)$, then
$$ \sum_{|I|=n} \mathbf{x}^I = \sum_i\frac{x_i^{n}}{\prod_{j\neq i}(1-\frac{x_j}{x_i})}. $$
With 1 variable, it gives
$$ x^n = x^n, $$
for two,
$$ \sum_{i+j = n} x^i y^j = \frac{x^{n+1}}{x-y} + \frac{y^{n+1}}{y-x}, $$
and for three, if gives
$$ \sum_{i+j+k=n} x^i y^j z^k= \frac{x^{n+2}}{(x-y)(x-z)} + \frac{y^{n+2}}{(y-x)(y-z)} + \frac{z^{n+2}}{(z-x)(z-y)}. $$
