please consider the following $n$-by-$n$ matrix:
$\\ A= \begin{bmatrix} 0! & 1! & 2! & \dots & n! \\ -1! & -2! & -3! & \dots & -(n+1)! \\ 2! & 3! & 4! & \dots& (n+2)! \\ \vdots & \vdots & \vdots & \dots & \vdots \\ \ (-1)^n n!&(-1)^n (n+1)!&(-1)^n (n+2)! &\dots & (-1)^n (2n)! \end{bmatrix}.$
I want to show that all the entries of $A^{-1}$ are bounded for finite values of $n$.
Can you help me?
My approach: I am seeking for an upper bound on the norm (1-norm preferably) of the inverse of matrix A. I am aware of the following theorem: $\Vert A^{-1}\Vert \le \frac{\Vert A \Vert ^{n-1}}{|det(A)|}$. However, following this strategy, I have to show that $det(A)$ is sufficiently far away from zero (I need a suitable lower bound for $det(A)$). I know that $A$ is non-singular, but it does not help.
Thanks.