Face-splitting product of two Vandermonde matrices: When is is invertible?
Sorry if I am missing something, but isn't it always singular in your special case? The first column is identically $1$, the second column is $(\beta_1,\ldots,\beta_{n^2})^T$, and the $n+1$'th column is $(\alpha_1,\ldots,\alpha_{n^2})^T$. If $\beta_i=\alpha_i-1$ for all $1\leq i\leq n^2$, then the $n+1$'th column is a linear combination of the first and second columns, so it is singular, right?
EDIT: in response to the amended question in the comments, I think it becomes nonsingular if and only if the $\alpha_i$ are distinct (note that one direction is obvious; if $\alpha_i=\alpha_j$ for some $i\neq j$, then clearly there are two equal rows).
To see the other direction, suppose that the $\alpha_i$ are distinct, but the matrix is singular. This means the $n+1$'th column is a linear combination of the previous $n$ columns. Defining $f_i(x)=x^i(x-1)^{n-i}$ for $i=0,\ldots,n$, this means there exists $\lambda_0,\ldots,\lambda_{n-1}\in \mathbb{F}$ not all zero such that for all $i=1,\ldots,n+1$,
\begin{equation}
f_{n}(\alpha_i) = \sum_{j=0}^{n-1} \lambda_j f_j(\alpha_i);
\end{equation}
as this holds at $n+1$ distinct points, and both sides have degree $n$ in $x$, this means that $f_n = \sum_{j=0}^{n-1} \lambda_j f_j$ as formal polynomials, as well as functions (the injectivity of the map between polynomials of degree at most $n$ and functions, when the field has at least $n+1$ elements, is essentially the nonsingularity of the Vandermonde matrix in the first place).
To see that this is not possible, note that evaluating both sides at $x=0$ shows that $\lambda_0 = 0$, for all other terms vanish, so this coefficient must be zero because $f_0(0)\neq 0$. Because all remaining terms are now divisible by $x$, we may safely divide through and keep this linear relationship, and then apply the same argument to deduce that $\lambda_1=0$, and so on. But this implies $f_n$ is formally the zero polynomial, which clearly does not hold. Therefore, the matrix has full column-rank and so is invertible.