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 & 2n-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.
1. **Indexing the Table:**
   - We index the rows from top to bottom as \(0, 1, \ldots, n-1\).
   - We index the columns from left to right as \(0, 1, \ldots, n-1\).
   - The value at row \(i\) and column \(j\) is given by \(a_{i,j} = ni + j\).

2. **Choosing \(n\) Numbers:**
   - We need to choose \(n\) numbers such that no two numbers are from the same row or the same column.
   - This implies we need to choose one number from each row and each column.

3. **Product of Chosen Numbers:**
   - To maximize the product, we need to consider the arrangement of indices.
   - We consider the product of the elements on the second diagonal, i.e., the elements \(a_{m, n-m-1}\) for \(m = 0, 1, \ldots, n-1\).

4. **Calculating the Product:**
   - The elements on the second diagonal are \(a_{0, n-1}, a_{1, n-2}, \ldots, a_{n-1, 0}\).
   - The value of each element \(a_{m, n-m-1}\) is given by:
     \[
     a_{m, n-m-1} = n \cdot m + (n - m - 1) = nm + n - m - 1
     \]
   - Therefore, the product of these elements is:
     \[
     \prod_{m=0}^{n-1} a_{m, n-m-1} = \prod_{m=0}^{n-1} (nm + n - m - 1)
     \]

5. **Simplifying the Product:**
   - Notice that \(a_{m, n-m-1}\) can be rewritten as:
     \[
     a_{m, n-m-1} = n(m + 1) - (m + 1) + (n - 1) = (n-1)(m+1) + (n-1)
     \]
   - This simplifies to:
     \[
     a_{m, n-m-1} = (n-1)(m+1) + (n-1) = (n-1)(m+1 + 1) = (n-1)(m+2)
     \]
   - Therefore, the product becomes:
     \[
     \prod_{m=0}^{n-1} a_{m, n-m-1} = \prod_{m=0}^{n-1} (n-1)(m+2)
     \]
   - This can be further simplified to:
     \[
     (n-1)^n \cdot \prod_{m=0}^{n-1} (m+2)
     \]
   - The product \(\prod_{m=0}^{n-1} (m+2)\) is equivalent to:
     \[
     2 \cdot 3 \cdot 4 \cdot \ldots \cdot (n+1) = \frac{(n+1)!}{1!} = (n+1)!
     \]

6. **Final Product:**
   - Therefore, the maximal value of the product is:
     \[
     (n-1)^n \cdot (n+1)!
     \]

The final answer is \( \boxed{ (n-1)^n \cdot (n+1)! } \)