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?
 A: 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
$$ \sum_{i+j+k=n} x^i y^j z^k= \frac{x^{n+2}}{(x-y)(x-z)} + \frac{y^{n+2}}{(y-x)(y-z)} + \frac{z^{n+2}}{(z-x)(z-y)}. $$
A: Found a simple proof: in induction:
Assume
$$\sum_{i_1 + i_2 + i_3 + ... +i_k = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = f[x_1,x_2,..,x_k]$$
For $$f(x) = x^{n + k - 1}$$
For every n. 
We'll show
$$\sum_{i_1 + i_2 + i_3 + ... +i_k + i_{k + 1} = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}{x_{k+1}}^{i_{k+1}} = g[x_1,x_2,..,x_k,x_{k+1}]$$
where
$$g(x) = x^{n + k}$$
Proof: by induction on k:
$$\sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = g[x_1,x_2,..,x_k]$$
$$\sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}{x_{k+1}}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = g[x_{k+1},x_2,..,x_k] = g[x_2,..,x_k,x_{k+1}]$$
Then by the definition of the divided difference:
$$ g[x_1,..,x_k,x_{k+1}] = \frac{g[x_1,..,x_k] - g[x_2,..,x_k,x_{k+1}]}{x_1 - x_{k+1}}$$
But then
$$g[x_1,..,x_k,x_{k+1}] =$$
$$ \frac{1}{x_1 - x_{k + 1}}\cdot\sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}\left({x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} - {x_{k+1}}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}\right) = \sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}\frac{{x_1}^{i_1} - {x_{k + 1}}^{i_1}}{x_1 - x_{k + 1}}
 \left({x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} \right)$$
$$ = \sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}\sum_{j_1 + j_2 = i_1 - 1}
 {x_1}^{j_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}{x_{k+1}}^{j_2} $$
$$= \sum_{i_1 + i_2 + i_3 + ... +i_k + i_{k + 1} = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}{x_{k+1}}^{i_{k+1}}$$
That's nice (and directly from the definition), but if someone has a more geometrical explanation for this I'll be glad to hear.
