Find all positive integers $n$ for which the number $1!+2!+3!+\cdots+n!$ is a perfect power of an integer.
To find all positive integers \( n \) for which the sum \( 1! + 2! + 3! + \cdots + n! \) is a perfect power of an integer, we will analyze the properties of factorials and modular arithmetic.

1. **Consider the sum modulo 10 for \( n \geq 4 \):**
   - For \( n \geq 4 \), the factorials \( 4!, 5!, 6!, \ldots \) all end in zero because they include the factor \( 10 \) (i.e., \( 4! = 24, 5! = 120, 6! = 720, \ldots \)).
   - Therefore, the sum \( 1! + 2! + 3! + \cdots + n! \) modulo 10 is determined by the first three terms: \( 1! + 2! + 3! \).

2. **Calculate \( 1! + 2! + 3! \) modulo 10:**
   \[
   1! = 1 \\
   2! = 2 \\
   3! = 6 \\
   1! + 2! + 3! = 1 + 2 + 6 = 9
   \]
   Thus, for \( n \geq 4 \):
   \[
   1! + 2! + 3! + \cdots + n! \equiv 9 \pmod{10}
   \]

3. **Check if 9 can be a perfect power modulo 10:**
   - The possible remainders of perfect squares modulo 10 are \( 0, 1, 4, 5, 6, 9 \).
   - The possible remainders of perfect cubes modulo 10 are \( 0, 1, 8, 9 \).
   - The possible remainders of perfect fourth powers modulo 10 are \( 0, 1, 5, 6 \).

   Since 9 is a possible remainder for both perfect squares and perfect cubes, we need to check if the sum \( 1! + 2! + 3! + \cdots + n! \) can be a perfect power for \( n \leq 3 \).

4. **Check \( n = 1, 2, 3 \):**
   - For \( n = 1 \):
     \[
     1! = 1
     \]
     1 is a perfect power (specifically, \( 1^k \) for any \( k \geq 1 \)).

   - For \( n = 2 \):
     \[
     1! + 2! = 1 + 2 = 3
     \]
     3 is not a perfect power.

   - For \( n = 3 \):
     \[
     1! + 2! + 3! = 1 + 2 + 6 = 9
     \]
     9 is a perfect power (specifically, \( 3^2 \)).

Therefore, the only values of \( n \) for which \( 1! + 2! + 3! + \cdots + n! \) is a perfect power are \( n = 1 \) and \( n = 3 \).

The final answer is \( \boxed{ n = 1 } \) and \( n = 3 \).