$\frac{n(n + 1)}{2}$ distinct numbers are arranged at random into $n$ rows. The first row has $1$ number, the second has $2$ numbers, the third has $3$ numbers and so on. Find the probability that the largest number in each row is smaller than the largest number in each row with more numbers.
To find the probability that the largest number in each row is smaller than the largest number in each subsequent row, we can use induction.

1. **Base Case: \( n = 1 \)**
   - For \( n = 1 \), we have only one row with one number. The probability that the largest number in this row is smaller than the largest number in the next row (which does not exist) is trivially 1.
   - Therefore, the base case holds.

2. **Inductive Step:**
   - Assume the formula holds for \( n = k \), i.e., the probability is \( \frac{2^k}{(k+1)!} \).
   - We need to show that the formula holds for \( n = k+1 \).

3. **Arrangement for \( n = k+1 \):**
   - For \( n = k+1 \), we have \( \frac{(k+1)(k+2)}{2} \) numbers arranged in \( k+1 \) rows.
   - The first row has 1 number, the second row has 2 numbers, ..., and the \( (k+1) \)-th row has \( k+1 \) numbers.

4. **Largest Element Placement:**
   - The largest element, \( \frac{(k+1)(k+2)}{2} \), must be in the \( (k+1) \)-th row to satisfy the condition.
   - The probability that the largest element is in the \( (k+1) \)-th row is:
     \[
     \frac{k+1}{\frac{(k+1)(k+2)}{2}} = \frac{2}{k+2}
     \]

5. **Relabeling Remaining Elements:**
   - After placing the largest element in the \( (k+1) \)-th row, we are left with \( \frac{k(k+1)}{2} \) elements to be arranged in \( k \) rows.
   - By the inductive hypothesis, the probability that these \( k \) rows satisfy the condition is \( \frac{2^k}{(k+1)!} \).

6. **Combining Probabilities:**
   - The total probability for \( n = k+1 \) is the product of the probability of placing the largest element correctly and the probability that the remaining elements satisfy the condition:
     \[
     \frac{2}{k+2} \times \frac{2^k}{(k+1)!} = \frac{2^{k+1}}{(k+2)(k+1)!} = \frac{2^{k+1}}{(k+1+1)!}
     \]

7. **Conclusion:**
   - By induction, the formula holds for all \( n \).

The final answer is \(\boxed{\frac{2^n}{(n+1)!}}\).