Let $n$ be a positive integer. Assume that $n$ numbers are to be chosen from the table

$$
\begin{array}{cccc}
0 & 1 & \cdots & n-1 \\
n & n+1 & \cdots & 2 n-1 \\
\vdots & \vdots & \ddots & \vdots \\
(n-1) n & (n-1) n+1 & \cdots & n^{2}-1
\end{array}
$$

with no two of them from the same row or the same column. Find the maximal value of the product of these $n$ numbers.
The product is $R(\sigma)=\prod_{i=0}^{n-1}(n i+\sigma(i))$, for some permutation $\sigma:\{0,1, \ldots, n-1\} \rightarrow\{0,1, \ldots, n-1\}$. Let $\sigma$ be such that $R(\sigma)$ is maximal. We may assume that all the multipliers $n i+\sigma(i)$ are positive, because otherwise the product is zero, that is the smallest possible.
Assume further that $\sigma(a)>\sigma(b)$ for some $a>b$. Let a permutation $\tau$ be defined by

$$
\tau(i)= \begin{cases}\sigma(b), & i=a \\ \sigma(a), & i=b \\ \sigma(i), & \text { otherwise }\end{cases}
$$

We have

$$
\frac{R(\tau)}{R(\sigma)}=\frac{(n a+\tau(a))(n b+\tau(b))}{(n a+\sigma(a))(n b+\sigma(b))}=\frac{(n a+\sigma(b))(n b+\sigma(a))}{(n a+\sigma(a))(n b+\sigma(b))}>1
$$

as
$(n a+\sigma(b))(n b+\sigma(a))-(n a+\sigma(a))(n b+\sigma(b))=n(a \sigma(a)+b \sigma(b)-a \sigma(b)-b \sigma(a))=n(a-b)(\sigma(a)-\sigma(b))>0$.
This is a contradiction with the maximality of $R(\sigma)$, hence, $\sigma$ has to satisfy $\sigma(a)<\sigma(b)$ for all $a<b$. Thus, $\sigma(i)=n-1-i$ for all $i$, and

$$
R(\sigma)=\prod_{i=0}^{n-1}(n i+n-1-i)=\prod_{i=0}^{n-1}(i+1)(n-1)=(n-1)^{n} n!
$$