Find all pairs $(p, n)$ with $n>p$, consisting of a positive integer $n$ and a prime $p$, such that $n^{n-p}$ is an $n$-th power of a positive integer.
1. **Claim and Initial Setup**:
   We claim that \((n, p) = (4, 2)\) is the only solution. Let \(n = p_1^{e_1} \cdots p_L^{e_L}\) for primes \(p_1 < \cdots < p_L\) and \(e_i \ge 1\). Note that \(n \mid e_i(n - p)\) for every \(i\).

2. **Case Analysis**:
   - **Case 1: \(p \nmid n\)**:
     If \(p \nmid n\), then \((n - p, n) = 1\), so \(n \mid e_i\) for all \(i\). This implies \(n \ge 2^n\), which is absurd for \(n > 1\). Therefore, \(p \mid n\).

   - **Case 2: \(p \mid n\)**:
     Set \(n = pk\) with \(k \ge 2\). We have that for some \(u\), \((pk)^{pk - p} = u^{pk}\), which simplifies to \((pk)^{k - 1} = u^k\). This implies that \(pk\) is a perfect \(k\)-th power. Set \(pk = \ell^k\). 

3. **Divisibility and Inequality**:
   Since \(p \mid \ell\), we can write \(\ell = pm\). Substituting, we get \(pk = (pm)^k\), which simplifies to \(k = p^{k - 1} m^k\). This implies \(p^{k - 1} \mid k\). Therefore, \(k \ge p^{k - 1}\).

4. **Checking for \(p = 2\)**:
   - If \(p = 2\), then \(k \ge 2^{k - 1}\). We check the values of \(k\):
     - For \(k = 1\), \(2 \ge 2^0 = 1\), which is true but \(k \ge 2\).
     - For \(k = 2\), \(2 \ge 2^1 = 2\), which is true.
     - For \(k = 3\), \(3 \ge 2^2 = 4\), which is false.
     - For \(k \ge 4\), \(k \ge 2^{k - 1}\) is false as \(2^{k - 1}\) grows exponentially.

   Thus, the only valid \(k\) for \(p = 2\) is \(k = 2\), giving \(n = pk = 2 \cdot 2 = 4\).

5. **Checking for \(p \ge 3\)**:
   - If \(p \ge 3\), then \(k \ge p^{k - 1}\). For \(p = 3\), \(k \ge 3^{k - 1}\):
     - For \(k = 2\), \(2 \ge 3^1 = 3\), which is false.
     - For \(k \ge 3\), \(k \ge 3^{k - 1}\) is false as \(3^{k - 1}\) grows exponentially.

   Therefore, there are no valid \(k\) for \(p \ge 3\).

Conclusion:
The only solution is \((n, p) = (4, 2)\).

The final answer is \( \boxed{ (4, 2) } \).