TAOCP 1.2.5 Exercise 15
The $(i,j)$ entry of the matrix is $i\times j$.
Section 1.2.5: Permutations and Factorials
Exercise 15. [**] [HM15] The permanent of a square matrix is defined by the same expansion as the determinant except that each term of the permanent is given a plus sign while the determinant alternates between plus and minus. Thus the permanent of
$$ \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} $$
is $aei+bfg+cdh+gec+hfa+idb$. What is the permanent of
$$ \begin{pmatrix} 1\times 1 & 1\times 2 & \cdots & 1\times n \ 2\times 1 & 2\times 2 & \cdots & 2\times n \ \vdots & \vdots & \ddots & \vdots \ n\times 1 & n\times 2 & \cdots & n\times n \end{pmatrix},? $$
Verified: yes
Solve time: 2m56s
The $(i,j)$ entry of the matrix is $i\times j$. Hence row $i$ is equal to $i$ times the first row:
$$ (i\times1,\ i\times2,\ \ldots,\ i\times n) = i(1,\ 2,\ \ldots,\ n). $$
Therefore, for every permutation $\pi$ of ${1,2,\ldots,n}$, the corresponding term in the permanent expansion is
$$ \prod_{i=1}^n (i\times \pi(i)) = \left(\prod_{i=1}^n i\right)\left(\prod_{i=1}^n \pi(i)\right). $$
Since $\pi$ is a permutation,
$$ \prod_{i=1}^n \pi(i)=1\cdot2\cdots n=n!, $$
and also
$$ \prod_{i=1}^n i=n!. $$
Thus every term of the permanent is equal to
$$ (n!)^2. $$
There are exactly $n!$ permutations, hence exactly $n!$ terms in the permanent expansion. Therefore the permanent is
$$ n!(n!)^2=(n!)^3. $$
Thus the value of the permanent is
$$ \boxed{(n!)^3}. $$