Example 7 A cube of a positive integer is a four-digit number, and the fourth power of this positive integer is a six-digit number. The 10 digits used here are exactly $0,1,2,3,4,5,6,7,8,9$, each appearing once, without repetition or omission. Try to find this positive integer.

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Let the positive integer be $n$. If $n$ is a single-digit number, then $n^{3}$ can be at most a three-digit number, and $n^{4}$ can be at most a four-digit number. Writing $n^{3}$ and $n^{4}$ together would certainly not use up 10 digits, so $n$ cannot be a single-digit number.

If $n$ is a three-digit number, even the smallest $100, n^{3}$ would be a seven-digit number, and $n^{4}$ would exceed seven digits. Writing $n^{3}$ and $n^{4}$ together would certainly exceed 10 digits, which does not meet the conditions given in the problem. Therefore, $n$ must be a two-digit number.

When $n$ is a two-digit number, $n^{4}$ must have at least one more digit than $n^{3}$. Therefore, $n^{3}$ cannot have more than four digits, and $n^{4}$ cannot have fewer than six digits. Since $22^{3}=10648$, which is already a five-digit number, $n$ must be less than 22; and since $17^{4}=83521$, which is only a five-digit number and not six, $n$ must be greater than 17. Therefore, $n$ can only be one of the numbers 18, 19, 20, or 21.

By calculation, it is not difficult to find that: $20^{3}=8000, 19^{4}=130321, 21^{4}=194481$, all of which have repeated digits, which does not meet the conditions of the problem and should be excluded. The only number left that might meet the conditions is 18. Since $18^{3}=5832, 18^{4}=104976$, and all 10 digits appear without repetition, the positive integer sought is 18.

Similarly, the answer to the following problem is also 18:
Problem: A positive integer, when multiplied by $1, 2, 3, 4, 5$ respectively, and the 5 products are written in sequence, exactly uses the 10 digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ without repetition or omission. Find this positive integer.