 Solution

Let $L_{ik}=1$ when $k\le i$ and $L_{ik}=0$ otherwise. The matrix product has entriesso $A_n=L{L}^T$. Both $L$ and its matrix transpose are triangular matrices with diagonal entries one. Multiplicativity of the determinant therefore gives $\det A_n=(\det L{)}^2=1$ for every $n\ge 1$. This determinant of the minimum-index matrix also follows by subtracting each preceding row from the next, working from the bottom upwards: the resulting matrix is upper triangular with unit diagonal.