Let matrix $$A=\begin{bmatrix} \binom{m}{k}&\binom{m}{k+1}&\cdots&\binom{m}{k+n-1}\\ \binom{m+1}{k}&\binom{m+1}{k+1}&\cdots&\binom{m+1}{k+n-1}\\ \cdots&\cdots&\cdots&\cdots\\ \binom{m+n-1}{k}&\binom{m+n-1}{k+1}&\cdots&\binom{m+n-1}{k+n-1} \end{bmatrix}$$ Find $\det{(A)}$
I know maybe use this well-known identity: $$\binom{m+i}{j-1}+\binom{m+i}{j}=\binom{m+i+1}{j}\Longrightarrow \binom{m+i+1}{j}-\binom{m+i}{j}=\binom{m+i}{j-1} $$ $$A\to\begin{bmatrix} \binom{m}{k}&\binom{m}{k+1}&\cdots&\binom{m}{k+n-1}\\ \binom{m}{k-1}&\binom{m}{k}&\cdots&\binom{m}{k+n-2}\\ \cdots&\cdots&\cdots&\cdots\\ \binom{m+n-2}{k-1}&\binom{m+n-2}{k}&\cdots&\binom{m+n-2}{k+n-2} \end{bmatrix}$$
I have used this identity many times(because if $k=0$, I can turn int $A$ into a diagonal matrix, and easily find this value $\det{(A)}=1$). But I can't turn into this matrix $A$ diagonal matrix.
PS: this problem is from book problem excise.see Linear algebra and matrix theory
Call the determinant $D(m,k)$ (I'll consider $n$ as fixed).
If $k>m$ the first row is zero, so $D(m,k)=0$ in this case.
Now assume $k\le m$. Applying the identity $$ \binom{a}{b}=\frac{a}{b}\binom{a-1}{b-1} $$ to each entry in the determinant, and taking out factors from rows and columns, we have $$ D(m,k)=\frac{m(m+1)(m+2)\dots(m+n-1)}{k(k+1)(k+2)\dots(k+n-1)} \, D(m-1,k-1) . $$ Repeating this, we work our way down to $$ D(m,k)=(\text{something}) \times D(m-k,0) , $$ and $D(m-k,0)=1$, as you already noticed. (A slick proof uses the Gessel–Viennot path-counting lemma; see pp. 221–222 in Martin Aigner's A Course in Enumeration).
Hence, $D(m,k)$ equals that "something", which I leave for you to work out. ;-)