Determinant of a Pascal Matrix, sort of

Answer : $n+1$

Explanation : It can be compute by using Pascal triangle formula ${n+1 \choose k+1} = {n \choose k} + {n \choose k+1}$.

First the determinant is the same by transforming each columns from $n+1$ to $2$ by it value minus the value of the previous column ($C_n$ become $C_n-C_{n-1}$). As ${n+1 \choose k+1}-{n \choose k+1} = {n \choose k}$ then

$$\begin{vmatrix} {1+1 \choose 1} & {1+2 \choose 1} & {1+3 \choose 1} & \dots & {1+n+1 \choose 1} \\ {2+1 \choose 2} & {2+2 \choose 2} & {2+3 \choose 2} & \dots & {2+n+1 \choose 2} \\ {3+1 \choose 3} &{3+2 \choose 3} & {3+3 \choose 3} & \dots & {3+n+1 \choose 3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {n+1+1 \choose n+1} & {n+1+2 \choose n+1} & {n+1+3 \choose n+1} & \dots & {n+1+n+1 \choose n+1} \end{vmatrix} = \begin{vmatrix} {1+1 \choose 1} & {1+1 \choose 0} & {1+2 \choose 0} & \dots & {1+n \choose 0} \\ {2+1 \choose 2} & {2+1 \choose 1} & {2+2 \choose 1} & \dots & {2+n \choose 1} \\ {3+1 \choose 3} &{3+1 \choose 2} & {3+2 \choose 2} & \dots & {3+n \choose 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {n+1+1 \choose n+1} & {n+1+1 \choose n} & {n+1+2 \choose n} & \dots & {n+1+n \choose n} \end{vmatrix} = \begin{vmatrix} {1+1 \choose 1} & 1 & 1 & \dots & 1 \\ {2+1 \choose 2} & {2+1 \choose 1} & {2+2 \choose 1} & \dots & {2+n \choose 1} \\ {3+1 \choose 3} &{3+1 \choose 2} & {3+2 \choose 2} & \dots & {3+n \choose 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {n+1+1 \choose n+1} & {n+1+1 \choose n} & {n+1+2 \choose n} & \dots & {n+1+n \choose n} \end{vmatrix}$$

Second the determinant is the same by transforming each rows from $n+1$ to $2$ by it value minus the value of the previous row ($L_n$ become $L_n-L_{n-1}$). As ${n+1 \choose k+1}-{n \choose k} = {n \choose k+1}$ then $$\begin{vmatrix} {1+1 \choose 1} & 1 & 1 & \dots & 1 \\ {2+1 \choose 2} & {2+1 \choose 1} & {2+2 \choose 1} & \dots & {2+n \choose 1} \\ {3+1 \choose 3} &{3+1 \choose 2} & {3+2 \choose 2} & \dots & {3+n \choose 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {n+1+1 \choose n+1} & {n+1+1 \choose n} & {n+1+2 \choose n} & \dots & {n+1+n \choose n} \end{vmatrix} = \begin{vmatrix} {1+1 \choose 1} & 1 & 1 & \dots & 1 \\ {2 \choose 2} & {1+1 \choose 1} & {1+2 \choose 1} & \dots & {1+n \choose 1} \\ {3 \choose 3} &{2+1 \choose 2} & {2+2 \choose 2} & \dots & {2+n \choose 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {n+1 \choose n+1} & {n+1 \choose n} & {n+2 \choose n} & \dots & {n+n \choose n} \end{vmatrix} = \begin{vmatrix} 2 & 1 & 1 & \dots & 1 \\ 1 & {1+1 \choose 1} & {1+2 \choose 1} & \dots & {1+n \choose 1} \\ 1 &{2+1 \choose 2} & {2+2 \choose 2} & \dots & {2+n \choose 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & {n+1 \choose n} & {n+2 \choose n} & \dots & {n+n \choose n} \end{vmatrix}$$

We can remark than this matrix from row $2$ to $n+1$ (and column $2$ to $n+1$) is the lower matrix of our initial matrix. So by doing the same two operations as above from row (and column) $3$ to $n+1$, I will get $$\begin{vmatrix} 2 & 1 & 0 & \dots & 0 \\ 1 & 2 & 1 & \dots & 1 \\ 0 & 1 & {1+1 \choose 1} & \dots & {1+(n-1) \choose 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & {(n-1)+1 \choose n-1} & \dots & {(n-1)+(n-1) \choose n-1} \end{vmatrix}$$. I do the same two operation several time and then get $$\begin{vmatrix} 2 & 1 & 0 & \dots & 0 \\ 1 & 2 & 1 & \dots & 1 \\ 0 & 1 & 2 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 2 \end{vmatrix}$$. It is the matrix with $2$ on the principal diagonal and $1$ on the $2$ secondary diagonals. This determinant can be determine by setting $$I_{n+1} = \begin{vmatrix} 2 & 1 & 0 & \dots & 0 \\ 1 & 2 & 1 & \dots & 1 \\ 0 & 1 & 2 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 2 \end{vmatrix} = 2I_n-I_{n-1}$$ with $I_0=1$ and $I_1=1$. This recurrent equation give $I_n = n+1$.