Determine all pairs $(n,p)$ of nonnegative integers such that [list] [*] $p$ is a prime, [*] $n<2p$, [*] $(p-1)^{n} + 1$ is divisible by $n^{p-1}$. [/list]
1. **Case 1: \( p = 2 \)**
   - Given \( p = 2 \), we have:
     \[
     (p - 1)^n + 1 = 1^n + 1 = 2
     \]
     - For \( n = 2 \), \( 2 \) is divisible by \( 2^{2-1} = 2 \).
     - Therefore, \( (n, p) = (2, 2) \) is a solution.

2. **Case 2: \( n = 1 \)**
   - For any prime \( p \), we have:
     \[
     (p - 1)^1 + 1 = p
     \]
     - Since \( p \) is divisible by \( 1^{p-1} = 1 \), \( (n, p) = (1, p) \) is a solution for any prime \( p \).

3. **Case 3: \( n = 2 \)**
   - We need \( (p - 1)^2 + 1 \) to be divisible by \( 2^{p-1} \).
   - Since \( (p - 1)^2 \) is odd, \( (p - 1)^2 + 1 \) is even, implying \( p \) must be even.
   - The only even prime is \( p = 2 \), which we already considered in Case 1.

4. **Case 4: \( p, n \geq 3 \)**
   - Since \( p \) is odd, \( (p - 1)^n + 1 \) is odd, implying \( n \) must be odd.
   - Let \( x \) be the greatest prime divisor of \( n \). Then:
     \[
     x \mid n \implies x \mid n^{p-1} \implies x \mid (p-1)^n + 1
     \]
     - Thus, \( (p-1)^n \equiv -1 \pmod{x} \).
   - Since \( x \mid n \), \( x \) and \( n \) are coprime, so there exist integers \( a \) and \( b \) such that:
     \[
     an = b(x-1) + 1
     \]
     - Since \( n \) is odd, \( x \) is odd, making \( b(x-1) \) odd, and thus \( b(x-1) + 1 \) is even, implying \( a \) is odd.
   - Therefore:
     \[
     (p-1)^{an} \equiv -1 \pmod{x}
     \]
     - By Fermat's Little Theorem, \( (p-1)^{x-1} \equiv 1 \pmod{x} \) since \( x \) and \( p-1 \) are coprime.
     - Thus:
       \[
       (p-1)^{b(x-1)} \cdot (p-1) \equiv -1 \pmod{x}
       \]
       \[
       (1)^b \cdot (p-1) \equiv -1 \pmod{x}
       \]
       \[
       p-1 \equiv -1 \pmod{x} \implies x \mid p
       \]
     - Since \( x \) is a prime divisor of \( p \), \( x = p \).
     - Given \( x \mid n \) and \( n < 2p \), we have \( n = p \).

5. **Verification for \( n = p \)**
   - We need \( p^{p-1} \mid (p-1)^p + 1 \).
   - Expanding \( (p-1)^p + 1 \):
     \[
     (p-1)^p + 1 = p^p - p^{p-1} \binom{p}{1} + p^{p-2} \binom{p}{2} - \cdots + (-1)^p
     \]
     - Simplifying, we get:
       \[
       p^{p-1} \mid p^2 (p^{p-2} - p^{p-3} \binom{p}{1} + \cdots + \binom{p}{p-2} + 1)
       \]
     - Since the expression in brackets is not divisible by \( p \), \( p^{p-1} \mid p^2 \) implies \( p-1 \leq 2 \), so \( p \leq 3 \).
     - Given \( p \geq 3 \), we have \( p = 3 \), and thus \( n = 3 \).

Conclusion:
- The solutions are \( (1, p) \) for any prime \( p \), \( (2, 2) \), and \( (3, 3) \).

The final answer is \( \boxed{ (1, p) } \) for any prime \( p \), \((2, 2)\), and \((3, 3)\).