Let $R=A[x_1,...,x_n]$ for some commutative ring $A$. I am trying to prove that any alternating polynomial in $R$ can be written in the form $$ \sum_{\lambda}c_{\lambda}a_{\lambda+\delta}\quad\quad(*) $$ where $\lambda=(\lambda_1,...,\lambda_n)$ is a non-decreasing multiindex, $\delta=\delta_n=(n-1,n-2,...,1,0)$ and $$ a_{\lambda+\delta}=\operatorname{det}(x_i^{\lambda_j+n-j}). $$
What I have so far. Let $f\in R$. For any $1\leq i<j\leq n$ we have $$ f(x_1,...,x_i,...,x_j,...,x_n)=-f(x_1,...,x_j,...,x_i,...,x_n) $$ which means that if $f(x)=0$ whenever $x_i=x_j$. Therefore $(x_i-x_j)\mid f$. Since this is true for all such $i,j$ we have actually proved that $$ a_\delta\mid f $$ where $a_\delta$ is the Vandermonde determinant. Now, write $f=a_\delta p$ for some $p\in R$. In order to prove that the last expression can be written in the form (*) we only need to prove that for any monomial $x^\lambda=x_1^{\lambda_1}...x_n^{\lambda_n}$ we have $$ x^{\lambda}a_{\delta}=a_{\delta+\lambda} $$ and this is evident form the multilinearity of the determinant.
Is this ok? Also, does this actually work for all commutative rings?
EDIT: thank to darij grinberg's comment I think I get it now.
First, $x^{\lambda}a_{\delta}=\operatorname{det}(x_i^{\lambda_i+n-j})\neq a_{\lambda+\delta}$ (note the subscript of $\lambda$ is $i$ as opposed to $j$). So, this does not help (at least not in an obvious way).
Here is a proof that I have compiled:
Let $f$ be an alternating polynomial in $n$ variables. Write $$ f=\sum_{\alpha} c_{\alpha}x^{\alpha} $$ where the sum is over all multiindices with pairwise distinct components. Now, $$ (\forall\alpha,\forall\sigma\in S_n)\colon\quad c_{\alpha}=\operatorname{sgn}(\sigma)c_{\sigma(\alpha)} $$ The orbits of the action of $S_n$ on $\alpha$'s can be parametrized by strictly decreasing multiindices, that is, of the form $\lambda+\delta$ as above. Now, let $C_\lambda=c_{\operatorname{sort}(\alpha)}$. Grouping all monomials of the given orbit and then factoring out the coefficient, we get $$ f=\sum_{\alpha} c_{\alpha}x^{\alpha}=\sum_{\lambda}\sum_{\sigma\in S_n}C_{\lambda}\operatorname{sgn}(\sigma)x^{\sigma(\lambda+\delta)}=\sum_{\lambda}C_{\lambda}\operatorname{det}(x_i^{\lambda_j+n-j})=\sum_{\lambda}C_{\lambda}a_{\lambda+\delta} $$