TITLE: How to find this determinant of the binomial coefficient $\det{(A)}$
QUESTION [8 upvotes]: 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

REPLY [6 votes]: 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. ;-)