I introduce some work I have done below, and the questions are at the end.
For some $n>1$ fixed, let $Q_n$ be the set (not vector space) of polynomials with integer coefficients of degree exactly $n$. Then, I claim the set:
$$B = \{(x-k)^n:k\in \mathbb{Z}\}$$
is an additive basis for $Q_n$, by which I mean that any element of $Q_n$ may be expressed as a finite linear combination (over $\mathbb{Q}$) of the elements of $B$.
Indeed, any element of $Q_n$ has $n+1$ degrees of freedom, and I claim that any set of $n+1$ distinct elements in $B$ is linearly independent. If:
$$S=\{(x-k_0)^n, ..., (x-k_{n})^n\}$$
($k_i \neq k_j, i \neq j$) is such a subset of $B$, then, the matrix of coefficients of $x$, after removing factors of (non-zero) binomial coefficients from each column, is:
\begin{pmatrix} 1 & k_0 & \cdots & k_0^{n}\\ 1 & k_1 & & k_1^n\\ \vdots & & \ddots & \vdots\\ 1 & k_n & \cdots & k_n^n \end{pmatrix}
i.e. a square Vandermonde matrix, which is invertible.
Questions
Is the reasoning above correct? If not, is it nevertheless true that $B$ forms an additive basis for $Q_n$ as claimed?
Given that $B$ is an additive basis, is there a convenient way to express an element of $Q_n$ as a finite linear combination of elements of $B$? Is there any way to perhaps introduce an inner product or do some combinatorial wizardry?
Although this question seems to be in the field of linear algebra, I hope to use an answer to (2) to help solve certain ODEs.