Is the the $n \times n$ matrix $A_{ij} = (ij + 1)^m$, $m \geq n$ invertible?

Consider the $n \times n$ symmetric matrix A, whose $ij$-th entry is defined by $A_{ij} = (ij + 1)^m$ and $m \geq n$. Is this matrix invertible?

Approaches I've tried:

• Numeric attempts to find a counterexample over a range of $n$ and $m$ have failed. These experiments did suggest that $A$ may be positive definite.
• A unisolvence theorem approach does not work, because the $n$ polynomials defining the rows or columns are of order $m$ and evaluated at $n \leq m$ points.

Solution 1:

$A$ is positive definite (and hence invertible) when $m+1\ge n$. Let $V$ be the $n\times(m+1)$ Vandermonde matrix $$ \pmatrix{1&1^1&1^2&\cdots&1^m\\ 1&2^1&2^2&\cdots&2^m\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ 1&n^1&n^2&\cdots&n^m} $$ From binomial expansion, we get $A=VDV^T$, where $D=\operatorname{diag}\left(\binom{m}{0},\binom{m}{1},\ldots,\binom{m}{m}\right)$. Since $D$ is positive definite and $V$ has full row rank (because $m+1\ge n$), $A$ is positive definite.