<s>[INST] Evaluation of a specific determinant. 
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.

$$A_2
\begin{pmatrix}  1&  1\\
  1&  2
\end{pmatrix};
A_3 = \begin{pmatrix}  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{pmatrix};
A_4 = \begin{pmatrix}  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{pmatrix};
A_5 = \begin{pmatrix}  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2\\
  1&  2&  3&  3&  3\\
  1&  2&  3&  4&  4\\
  1&  2&  3&  4&  5
\end{pmatrix}$$
$$A_6 = \begin{pmatrix}  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4\\
  1&  2&  3&  4&  5&  5\\
  1&  2&  3&  4&  5&  6
\end{pmatrix};
A_7 = \begin{pmatrix}  1&  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4&  4\\
  1&  2&  3&  4&  5&  5&  5\\
  1&  2&  3&  4&  5&  6&  6\\
  1&  2&  3&  4&  5&  6&  7
\end{pmatrix}
$$
 [/INST] If you expand along the bottom row, then most of the $(n-1)\times(n-1)$ subdeterminants are $0$ because the $(n-1)\times(n-1)$ submatrices have their two right-most columns identical.
So expansion along the bottom row leaves only the last two terms: $$(-1)(n-1)\det(A_{n-1})+n\det(A_{n-1})$$ which is just $\det(A_{n-1})$. So inductively, $\det(A_{n})=\det(A_{n-1})=\cdots=\det(A_{1})=1$
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