Determinant of a Pascal Matrix, sort of Let $A_{n}$ be the $(n+1) \times(n+1)$ matrix with coefficients
$$
a_{i j}={i+j \choose i}
$$
(binomial coefficients), where the rows and columns are indexed by the numbers from
0 to $n$ are indexed.
Now I want to determine the Determinant and with the first 5 matrices i found out that it is $n+1$ if i did not make a mistake.
The Matrix looks like this:
$$
\left(\begin{matrix}
{1+1 \choose 1} & {1+2 \choose 1} & {1+3 \choose 1} & \dots & {1+n+1 \choose 1} \\
{2+1 \choose 2} & {2+2 \choose 2} & {2+3 \choose 2} & \dots & {2+n+1 \choose 2} \\
{3+1 \choose 3} &{3+2 \choose 3} & {3+3 \choose 3} & \dots & {3+n+1 \choose 3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
{n+1+1 \choose n+1} & {n+1+2 \choose n+1} & {n+1+3 \choose n+1} & \dots & {n+1+n+1 \choose n+1}
\end{matrix}\right)
$$
The Problem is to bring this Matrix into an upper or lower triangle matrix. If anyone has hints or ideas that can help, please help, thanks in advance. Maybe the approach is not even good. If I make progress at all i will update this question.
 A: Let's consider the version where the columns are indexed from $0$ to $n$. Consider for example
$$ A_3 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{bmatrix}. $$
Note that each element in the matrix (except the elements in the $0$th row and column) is the sum of the element above it and the element to the left of it. This is the "Pascal Triangle" property of the matrix.
How can we reduce $A_4$ to an upper triangular matrix? By the Pascal Triangle property, if we replace row $R_i$ by $R_i - R_{i-1}$, the resulting row is just $R_i$, shifted to the right (with the first element padded by zero and the last element dropped). So for example,
$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ \color{pink}{1} & \color{\pink}{2} & \color{\pink}{3} & 4 \\ \color{red}{1} & \color{red}{3} & \color{red}{6} & 10 \\ \color{blue}{1} & \color{blue}{4} & \color{blue}{10} & 20 \end{bmatrix} \xrightarrow{R_4 = R_4 - R_3} 
\begin{bmatrix} 1 & 1 & 1 & 1 \\ \color{pink}{1} & \color{pink}{2} & \color{pink}{3} & 4 \\ \color{red}{1} & \color{red}{3} & \color{red}{6} & 10 \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix} \xrightarrow{R_3 = R_3 - R_2}  
\begin{bmatrix} 1 & 1 & 1 & 1 \\ \color{pink}{1} & \color{pink}{2} & \color{pink}{3} & 4 \\ 0 & \color{red}{1} & \color{red}{3} & \color{red}{6} \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix} \xrightarrow{R_2 = R_2 - R_1}
\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\ 0 & \color{red}{1} & \color{red}{3} & \color{red}{6} \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix}. $$
Now the corner $3 \times 3$ matrix again satisfies the "Pascal Triangle" property so you can repeat this process and get
$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\ 0 & \color{red}{1} & \color{red}{3} & \color{red}{6} \\ 0 & \color{blue}{1} & \color{blue}{4} & \color{blue}{10} \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 \\
 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\
 0 & 0 & \color{red}{1} & \color{red}{3} \\
 0 & 0 & \color{blue}{1} & \color{blue}{4} \end{bmatrix}. $$
Repeating this process once again for the corner $2 \times 2$ matrix, we get
$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\
 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\
 0 & 0 & \color{red}{1} & \color{red}{3} \\
 0 & 0 & \color{blue}{1} & \color{blue}{4} \end{bmatrix} \rightarrow 
\begin{bmatrix} 1 & 1 & 1 & 1 \\
 0 & \color{pink}{1} & \color{pink}{2} & \color{pink}{3} \\
 0 & 0 & \color{red}{1} & \color{red}{3} \\
 0 & 0 & 0 & \color{blue}{1} \end{bmatrix} $$
which is an upper triangular matrix with $1$'s on the diagonal so the determinant of the matrix is one.
I'll leave you the details of generalizing this argument to the $(n+1)\times(n+1)$ case and how one can use it to calculate the shifted determinant.
