Let $n$ be a positive integer. Consider an $n\times n$ grid of unit squares. How many ways are there to partition the horizontal and vertical unit segments of the grid into $n(n + 1)$ pairs so that the following properties are satisfied?

(i) Each pair consists of a horizontal segment and a vertical segment that share a common endpoint, and no segment is in more than one pair.

(ii) No two pairs of the partition contain four segments that all share the same endpoint.

(Pictured below is an example of a valid partition for $n = 2$.)

[asy]
import graph; size(2.6cm); 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
draw((-3,4)--(-3,2)); 
draw((-3,4)--(-1,4)); 
draw((-1,4)--(-1,2)); 
draw((-3,2)--(-1,2)); 
draw((-3,3)--(-1,3)); 
draw((-2,4)--(-2,2)); 
draw((-2.8,4)--(-2,4), linewidth(2)); 
draw((-3,3.8)--(-3,3), linewidth(2)); 
draw((-1.8,4)--(-1,4), linewidth(2)); 
draw((-2,4)--(-2,3.2), linewidth(2)); 
draw((-3,3)--(-2.2,3), linewidth(2)); 
draw((-3,2.8)--(-3,2), linewidth(2)); 
draw((-3,2)--(-2.2,2), linewidth(2)); 
draw((-2,3)--(-2,2.2), linewidth(2)); 
draw((-1,2)--(-1.8,2), linewidth(2)); 
draw((-1,4)--(-1,3.2), linewidth(2)); 
draw((-2,3)--(-1.2,3), linewidth(2)); 
draw((-1,2.8)--(-1,2), linewidth(2)); 
dot((-3,2),dotstyle); 
dot((-1,4),dotstyle); 
dot((-1,2),dotstyle); 
dot((-3,3),dotstyle); 
dot((-2,4),dotstyle); 
dot((-2,3),dotstyle);[/asy]
1. **Define the Problem and Key Terms:**
   - We are given an \( n \times n \) grid of unit squares.
   - We need to partition the horizontal and vertical unit segments into \( n(n + 1) \) pairs.
   - Each pair consists of a horizontal segment and a vertical segment that share a common endpoint.
   - No segment is in more than one pair.
   - No two pairs of the partition contain four segments that all share the same endpoint.

2. **Identify the Total Number of Vertices:**
   - The grid has \( (n+1)^2 \) vertices.

3. **Isolated Vertices:**
   - Define an isolated vertex as one that is not a common endpoint of any pair.
   - There must be exactly \( n+1 \) isolated vertices because there are \( n(n+1) \) pairs and \( (n+1)^2 \) vertices.

4. **Occupied Rows and Columns:**
   - A row or column is occupied if every edge between two consecutive vertices in that row or column has been placed in a pair.
   - No two isolated vertices can share the same row or column.

5. **Orientation of Pairs:**
   - Pairs are similarly oriented if their corner lies on the same side of the horizontal edge.
   - Pairs are oppositely oriented otherwise.

6. **Necessity of Isolated Vertices:**
   - Since there are \( (n+1)^2 \) vertices and \( n(n+1) \) pairs, there must be \( n+1 \) isolated vertices.
   - If a vertex is isolated in a row, the pairs containing every edge in that row in one direction of the isolated vertex must be similarly oriented and oppositely oriented compared to the pairs containing the edges in the other direction of the isolated vertex.
   - This implies that every other vertex in the row becomes a corner, and there is at most one isolated vertex per row or column.

7. **Sufficiency and Uniqueness:**
   - Given \( n+1 \) marked isolated vertices, we can reconstruct the partition row-by-row uniquely.
   - Start with the bottom row, where the vertical edge of every partition containing a horizontal edge from the bottom row must be oriented up relative to the horizontal edge.
   - For each subsequent row, begin with the horizontal edge(s) neighboring the isolated vertex in that row.
   - If the downwards edge attached to the left neighbor of the isolated vertex is vacant, the pair containing that horizontal edge must contain that vertical edge as well.
   - If the vertical edge is occupied, the only option for the partition is to choose the upwards edge.

8. **Existence of Construction:**
   - The construction is always possible until the last row.
   - The number of gaps (vacant vertical edges) between rows \( i \) and \( i+1 \) equals precisely \( i \).
   - When reaching the final row, all horizontal edges in row \( n+1 \) and all but one vertical edge (under the final isolated vertex) are vacant.
   - We can always construct partitions downwards for the final row.

9. **Conclusion:**
   - There is a bijection between a valid coloring and a selection of \( n+1 \) isolated vertices with respect to the conditions.
   - The number of ways to choose \( n+1 \) isolated vertices such that no two share a row or column is \( (n+1)! \).

The final answer is \( \boxed{ (n+1)! } \).