Q:

basis for space of polynomials

In the proof of Chevalley's theorem on invariants in the polynomials on a semisimple Lie algebra, one uses the following general fact: the polynomials $(\sum m_ix_i)^n$ with $m_i$ positive integers span the homogeneous polynomials of degree $n$ in the variables $x_i$ with coefficients in a field of characteristic zero. Does anyone know a proof of this fact, ideally without painful combinatorics?

A:

New version of the answer
Let $K$ be a field of characteristic $0$, and let $S^m$ be the space of homogeneous degree $m$ polynomials in $x_1,\dots,x_n$ with coefficients in $K$. Identify $S^1$ to $K^n$ in the obvious way. Let $Z$ be a Zariski dense subset of $K^n$. 
We claim that the $m$-powers of elements of $Z$ generate $S^m$ over $K$. 
Let $M$ be the set of length $m$ multi-indices $\alpha$ in $\mathbb N^n$, and equip $S^m$ with the symmetric bilinear form $(f,g)\mapsto\langle f\ |\ g\rangle$ for which the monomials $x^\alpha$, $\alpha\in M$, form an orthonormal basis. 
Let $f\in S^m$ be orthogonal to $a^m$ for all $a$ in $Z$. It suffices to show $f=0$. 
For $\alpha$ in $M$, define $\binom{m}{\alpha}$ by the expansion 
$$
(x_1+\cdots+x_n)^m=\sum_{\alpha\in M}\ \binom{m}{\alpha}\ x^\alpha 
$$ 
in $\mathbb Z[x_1,\dots,x_n]$. The $\binom{m}{\alpha}$ are then positive integers; in particular, they are nonzero in $K$ (because $K$ is of characteristic $0$). 
In $K[x_1,\dots,x_n]$ we have, for $a$ in $K^n$,
$$
(a_1\,x_1+\cdots+a_n\,x_n)^m=\sum_{\alpha\in M}\ \binom{m}{\alpha}\ a^\alpha\ x^\alpha, 
$$ 
and thus 
$$
\Big\langle(a_1\,x_1+\cdots+a_1\,x_n)^m\ \Big|\ x^\alpha\Big\rangle=\binom{m}{\alpha}\ a^\alpha, 
$$ 
which (in view of the above identification $S^1=K^n$) we can write 
$$
\langle a^m\ |\ x^\alpha\rangle=\binom{m}{\alpha}\ a^\alpha. 
$$ 
Now, if $f=\sum b_\alpha\, x^\alpha$ is orthogonal to $a^m$ for all $a$ in $Z$, we get 
$$
\sum_{\alpha\in M}\ \binom{m}{\alpha}\ b_\alpha\ a^\alpha=0 
$$ 
for all $a$ in $Z$, which implies indeed $f=0$. 
Old version of the answer
Let $K$ be a field of characteristic $0$, and let $S^m$ be the space of homogeneous degree $m$ polynomials in $x_1,\dots,x_n$ with coefficients on $K$. Identify $S^1$ to $K^n$ in the obvious way. Let $Z$ be a Zariski dense subset of $K^n$. 
We claim that the $m$-powers of elements of $Z$ generate $S^m$ over $K$. 
Put 
$$
\partial_i:=\frac{\partial}{\partial x_i}\quad,\quad\partial:=(\partial_1,\dots,\partial_n). 
$$ 
The formula 
$$
(f\ |\ g):=f(\partial)\,g(x) 
$$ 
defines a non-degenerate symmetric bilinear form on $S^m$. 
Let $\alpha$ be an $n$-multi-index of length $m$. The assumption that $K$ is characteristic $0$ implies, by a straightforward argument, the existence of a nonzero scalar $s(\alpha)$ such that 
$$
(a^m\ |\ x^\alpha)=s(\alpha)\,a^\alpha 
$$ 
for all $a$ in $K^n$. [We're using standard multi-index notation.] 
Let $f\in S^m$ be orthogonal to $a^m$ for all $a$ in $Z$. By the above observation, $f=0$. This proves the claim.

