$ A_=\begin{pmatrix} 1&(-2)&0&0&\cdots&0&0&0\\0&2&(-3)&0&\cdots&0&0&0\\0&0&3&(-4)&\cdots&0&0&0\\0&0&0&4&\cdots&0&0&0\\\vdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\0&0&0&0&\cdots&(n-2)&(-(n-1))&0\\0&0&0&0&\cdots&0&(n-1)&(-n)\\1&1&1&1&\cdots&1&1&(1+n) \\ \end{pmatrix} $
I know that I have to put in $n=1$ and few other firsts but i don't know what to do with those few last rows...and I know I have to go look for a certain sequence. Help would be very appreciated.
For $n>=2$ the determinant is given by : $$(n+1)(n-1)! + (n!\sum_{i=1}^{n-2}\frac{1}{i})+n(n-2)!$$ The idea is to convert it into block matrices.
Step1. Add to every row $i\in1..n-2$, rows $(i+1)$ to $(n-1)$. For example$$Row1=Row1+Row2+...Row(n-1)$$ $$Row2=Row2+Row3+...Row(n-1)$$ $$...$$ $$Row(n-2)=Row(n-2)+Row(n-1)$$ Your matrix will look something like this :
$ A_=\begin{pmatrix} 1&0&0&0&\cdots&0&0&-n\\0&2&0&0&\cdots&0&0&-n\\0&0&3&0&\cdots&0&0&-n\\0&0&0&4&\cdots&0&0&-n\\\vdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\0&0&0&0&\cdots&(n-2)&0&-n\\0&0&0&0&\cdots&0&(n-1)&(-n)\\1&1&1&1&\cdots&1&1&(1+n) \\ \end{pmatrix} $
Step2. Now to convert this into block matrix we have to first remove $-n$ from last column from $i=1..n-2$. For that apply the following operation : $$col(n)=col(n)+\sum_{j=1}^{n-2}\frac{n*col(j)}{j}$$
matrix will be transformed to this :
$ A_=\begin{pmatrix} 1&0&0&0&\cdots&0&0&0\\0&2&0&0&\cdots&0&0&0\\0&0&3&0&\cdots&0&0&0\\0&0&0&4&\cdots&0&0&0\\\vdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\0&0&0&0&\cdots&(n-2)&0&0\\0&0&0&0&\cdots&0&(n-1)&(-n)\\1&1&1&1&\cdots&1&1&(1+n)+\sum_{i=1}^{n-2}\frac{n}{i}) \\ \end{pmatrix} $
Step3. Now remove 1's from the last row from $j=1..n-2$ by applying following operation : $$Row(n)=Row(n)-\sum_{i=1}^{n-2}\frac{Row(i)}{i}$$
Notice that this won't change anything else. Matrix will look like this : $ A_=\begin{pmatrix} 1&0&0&0&\cdots&0&0&0\\0&2&0&0&\cdots&0&0&0\\0&0&3&0&\cdots&0&0&0\\0&0&0&4&\cdots&0&0&0\\\vdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\0&0&0&0&\cdots&(n-2)&0&0\\0&0&0&0&\cdots&0&(n-1)&(-n)\\0&0&0&0&\cdots&0&1&(1+n)+\sum_{i=1}^{n-2}\frac{n}{i}) \\ \end{pmatrix} $
Now calculate the determinant of each block and multiply them to get your determinant or make 0 at $A_{n,n-1}$ and your matrix will transform to upper triangular whose determinant is just product of diagonal entries. The answer will be what I mentioned in the starting.