I am interested whether the $n\times n$ matrix $A=\{a_{ij}\}$ with
$$
a_{ij}
=
\begin{cases}
\dfrac{i!}{(i-j+1)!},
\;j\le i,\\[2mm]
0,\;\text{otherwise}
\end{cases}
$$
has a known inverse? The inverse itself would be of interest as well. I wrote a simple Mathematica script to see what's going on here, and here's what I got say for $n=15$:
I can see that the diagonal elements are $1/i!$, and the elements along the first subdiagonal are $-1/(2 i!)$, and along the second subdiagonal are $1/(12 i!)$. Not sure about the other elements.
By observing your matrix seriously, I can find that the general formula of $(b_{ij}) = A^{-1}$ is given by $$ b_{ij} = \begin{cases} \frac{B_{i-j}}{j!(i-j)!} & i\geq j \\ 0 & \text{otherwise}\end{cases} $$ where $B_{m}$ is an $m$-th Bernoulli number. This follows from the formula in the recursive definition.