For each positive integer $n$ let $A_n$ be the $n \times n$ matrix such that its $a_{ij}$
 entry is equal to ${i+j-2 \choose j-1}$ for all $1\leq i,j \leq n.$ Find the determinant of $A_n$.

1. We start by examining the given matrix \( A_n \) for small values of \( n \) to identify any patterns. For \( n = 2 \), \( n = 3 \), and \( n = 4 \), the matrices are:
   \[
   A_2 = \begin{bmatrix}
   1 & 1 \\
   1 & 2
   \end{bmatrix}
   \]
   \[
   A_3 = \begin{bmatrix}
   1 & 1 & 1 \\
   1 & 2 & 3 \\
   1 & 3 & 6
   \end{bmatrix}
   \]
   \[
   A_4 = \begin{bmatrix}
   1 & 1 & 1 & 1 \\
   1 & 2 & 3 & 4 \\
   1 & 3 & 6 & 10 \\
   1 & 4 & 10 & 20
   \end{bmatrix}
   \]

2. We observe that the entries of the matrix \( A_n \) are given by the binomial coefficients:
   \[
   a_{ij} = \binom{i+j-2}{j-1}
   \]
   This is a well-known binomial coefficient identity.

3. We notice a pattern in the entries of the matrices. Specifically, we observe that:
   \[
   a_{ij} + a_{i+1,j-1} = a_{i+1,j}
   \]
   This is Pascal's Identity:
   \[
   \binom{i+j-2}{j-1} + \binom{i+j-2}{j-2} = \binom{i+j-1}{j-1}
   \]

4. To simplify the determinant calculation, we will perform row operations to transform the matrix into an upper triangular form. We start by subtracting the \( i \)-th row from the \( (i+1) \)-th row, starting from \( i = n-1 \) and moving upwards.

5. For \( n = 4 \), the row operations are as follows:
   \[
   \begin{bmatrix}
   1 & 1 & 1 & 1 \\
   1 & 2 & 3 & 4 \\
   1 & 3 & 6 & 10 \\
   1 & 4 & 10 & 20
   \end{bmatrix}
   \rightarrow
   \begin{bmatrix}
   1 & 1 & 1 & 1 \\
   1 & 2 & 3 & 4 \\
   1 & 3 & 6 & 10 \\
   0 & 1 & 4 & 10
   \end{bmatrix}
   \rightarrow
   \begin{bmatrix}
   1 & 1 & 1 & 1 \\
   1 & 2 & 3 & 4 \\
   0 & 1 & 3 & 6 \\
   0 & 1 & 4 & 10
   \end{bmatrix}
   \rightarrow
   \begin{bmatrix}
   1 & 1 & 1 & 1 \\
   0 & 1 & 2 & 3 \\
   0 & 1 & 3 & 6 \\
   0 & 1 & 4 & 10
   \end{bmatrix}
   \]

6. Continuing this process, we can transform the matrix into an upper triangular form with 1's on the diagonal and 0's below the diagonal. This is justified by the Pascal's Identity pattern found earlier.

7. Once the matrix is in upper triangular form, the determinant of the matrix is the product of the diagonal entries. Since all diagonal entries are 1, the determinant is:
   \[
   \boxed{1}
   \]