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 $i,j\in \{1, \ldots, n\}$.

$$A_2
\begin{pmatrix}  1&  1\\
  1&  2
\end{pmatrix};
A_3 = \begin{pmatrix}  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{pmatrix};
A_4 = \begin{pmatrix}  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{pmatrix};
A_5 = \begin{pmatrix}  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2\\
  1&  2&  3&  3&  3\\
  1&  2&  3&  4&  4\\
  1&  2&  3&  4&  5
\end{pmatrix}$$
$$A_6 = \begin{pmatrix}  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4\\
  1&  2&  3&  4&  5&  5\\
  1&  2&  3&  4&  5&  6
\end{pmatrix};
A_7 = \begin{pmatrix}  1&  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4&  4\\
  1&  2&  3&  4&  5&  5&  5\\
  1&  2&  3&  4&  5&  6&  6\\
  1&  2&  3&  4&  5&  6&  7
\end{pmatrix}
$$
 A: $$A_2 = \begin{bmatrix}1 & 1\\ 1 & 2 \end{bmatrix} = \begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$$
$$A_3 = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\1 & 2 & 3 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}$$
$$A_4 = \begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{bmatrix}$$
Can you see the pattern here? Prove this is the case in general.
Use this, along with the fact that $\det(XY) = \det(X) \cdot \det(Y)$ and determinant of a triangular matrix is the product of the diagonal entries to conclude the answer.
EDIT
Note that $$A_{ij} = \min(i,j) = \sum_{k=1}^{\min(i,j)} 1 = \sum_{k=1}^{n} \mathbb{I}_{k \leq i}\mathbb{I}_{k \leq j} = \sum_{k=1}^n B_{ik} C_{kj}$$
where $B_{ik} = \begin{cases} 1 & k \leq i\\ 0 & \text{else}\end{cases}$ and $C_{kj} = \begin{cases} 1 & k \leq j\\ 0 & \text{else}\end{cases}$.
This should enable you to get the decomposition.
A: 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, $\det(A_{n})=\det(A_{n-1})=\cdots=\det(A_{1})=1$
A: If you consider the matrix 
$$
B_n=\begin{bmatrix}
1&0&\cdots&0\\
1&1&0&\cdots\\
&&\cdots\\
1&1&\cdots &1
\end{bmatrix}
$$
(i.e. the $i,j$ entry of $B_n$ is $1$ is $i\geq j$ and $0$ otherwise), then the $k^{\rm t h}$ column of $A_n$ is obtained by adding the first $k$ columns of $B_n$. So
$$
\det A_n=\det B_n=1.
$$
A: Recall that adding a multiple or subtracting a multiple of one row does not change the value of the determinant, see, for example ProofWiki.
Using this fact and Laplace expansion you get
$$|A_4|=
\begin{vmatrix}
  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{vmatrix}=
\begin{vmatrix}
  1&  1&  1&  1\\
  0&  1&  1&  1\\
  0&  1&  2&  2\\
  0&  1&  2&  3
\end{vmatrix}=
\begin{vmatrix}
  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{vmatrix}=|A_3|.$$
(We have subtracted the first row from the other rows.)
In the same way you get $|A_{n+1}|=|A_n|$.
