Given some variables $x_1, x_2, \dots, x_n$, the Vandermonde determinant is given by $$V_n(x_1,\dots,x_n):=\det(x_j^{i-1})_{i,j=1}^n=\prod_{i<j}(x_j-x_i).$$ One can take as special cases: $x_j=j$ or $x_j=q^j$.

I was playing around with the following variation $A_n=((i+j-1)^{i-1})_{i,j=1}^n$. It turns out that (numerically) $\det(A_n)=\det(j^{i-1})_{i,j=1}^n$.

In addition, if $B_n=(q^{(i+j-1)(i-1)})_{i,j=1}^n$ then $\det(B_n)$ and $\det(q^{j(i-1)})_{i,j=1}^n$ differ only by a factor of a power of $q$, that is, $q^c$ for some $c$.

On the other hand, if $C_n=(x_{i+j-1}^{i-1})_{i,j=1}^n$ then (unfortunately) $\det(C_n)$ and $V_n$ are vastly different; in fact, $\det(C_n)$ does not factor as $V_n$.

QUESTION 1.Is this true? $\det(A_n)=\det(j^{i-1})_{i,j=1}^n$.

QUESTION 2.What worked right for $A_n$ andd $B_n$ and failed for $C_n$?