Let A be an $n\times n$ square matrix with a$_{ij} = (i+j-1)!$. Compute $\text{det}(A)$.
I've tried to factorise from each column the elements of the first row and then performed some row operations to generate 0's to use Laplace expansion but with no success.
I'm thinking that there might be some kind of recurrence relation for the determinant in terms of lower-sized determinants with the same structure.
Denote the given matrix by $A_n$. From $A_n$, divide row $i$ by $i!$ for all $i$. Also divide column $j$ by $(j-1)!$ for all $j$. We obtain matrix $B_n$ whose entry at $(i,j)$ is $ \frac{(i+j-1)!}{i!(j-1)!}={i+j-1\choose j-1}$. $$\det(A_n)=\det(B_n)\left(\prod_{i=1}^ni!\right)\left(\prod_{j=1}^n(j-1)!\right)$$
$$B_5=C_5=\pmatrix{ {1\choose 0},&{2\choose 1},&{3\choose 2},&{4\choose 3}\\ {2\choose 0},&{3\choose 1},&{4\choose 2},&{5\choose 3}\\ {3\choose 0},&{4\choose 1},&{5\choose 2},&{6\choose 3}\\ {4\choose 0},&{5\choose 1},&{6\choose 2},&{7\choose 3}\\ }$$ For integer $t$ between $1$ and $n$, let matrix $C_{t}$ be the $t\times t$ matrix whose entry at $(i,j)$ is $i+j-1+n-t\choose j-1$. In particular, $C_n=B_n$.
From $C_t$ where $t\ge2$, subtract each row from the row above it, starting at the bottom bow and ending at the second row. We obtain matrix $D_t$ whose entry at $(i,j)$, $i\ge2$ is $${i+j-1+n-t\choose j-1}-{i-1+j-1+n-t\choose j-1}={i-1+j-1+n-t\choose j-2}.$$ $$D_5=\pmatrix{ {1\choose 0},&{2\choose 1},&{3\choose 2},&{4\choose 3}\\ 0,&{2\choose 0},&{3\choose 1},&{4\choose 2}\\ 0,&{3\choose 0},&{4\choose 1},&{5\choose 2}\\ 0,&{4\choose 0},&{5\choose 1},&{6\choose 2}\\ }$$ So for $D_t$,
Hence $\det(C_t)=\det(D_t)=\det(C_{t-1})$, for all $2\le t\le n$. That implies $\det(C_n)=\det(C_1)=1$. So $\det(B_n)=\det(C_n)=1$.
$$\det(A_n)=\left(\prod_{i=1}^ni!\right)\left(\prod_{j=1}^n(j-1)!\right)=n!(\prod_{i=1}^{n-1}j!)^2$$