= Solution
Let $L_{ik}=1$ when $k\leq i$ and $L_{ik}=0$ otherwise. The <matrix product> has entries
$$
(LL^T)_{ij}=\sum_{k=1}^n L_{ik}L_{jk}=\min(i,j),
$$
so $A_n=LL^T$. Both $L$ and its <matrix transpose> are <triangular matrices> with diagonal entries one. Multiplicativity of the <determinant> therefore gives \b[$\det A_n=(\det L)^2=1$ for every $n\geq1$]. 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.
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