Determine all prime numbers of the form $1 + 2^p + 3^p +...+ p^p$ where $p$ is a prime number.
1. **Case for \( p = 2 \):**
   \[
   1 + 2^2 = 1 + 4 = 5
   \]
   Since 5 is a prime number, \( p = 2 \) is a valid solution.

2. **Case for \( p > 2 \):**
   - Since \( p \) is a prime number greater than 2, \( p \) must be odd.
   - Consider the sum \( S = 1 + 2^p + 3^p + \cdots + p^p \).

3. **Modulo \( p \) Analysis:**
   - For any integer \( a \), \( a^p \equiv a \pmod{p} \) by Fermat's Little Theorem.
   - Therefore, \( a^p + (-a)^p \equiv a + (-a) \equiv 0 \pmod{p} \) for \( a \neq 0 \).

4. **Grouping Terms:**
   - Group the terms in pairs:
     \[
     (1^p + (p-1)^p), (2^p + (p-2)^p), \ldots, \left(\left(\frac{p-1}{2}\right)^p + \left(\frac{p+1}{2}\right)^p\right)
     \]
   - Each pair sums to 0 modulo \( p \):
     \[
     1^p + (p-1)^p \equiv 1 + (-1) \equiv 0 \pmod{p}
     \]
     \[
     2^p + (p-2)^p \equiv 2 + (-2) \equiv 0 \pmod{p}
     \]
     \[
     \vdots
     \]
     \[
     \left(\frac{p-1}{2}\right)^p + \left(\frac{p+1}{2}\right)^p \equiv \left(\frac{p-1}{2}\right) + \left(-\frac{p-1}{2}\right) \equiv 0 \pmod{p}
     \]

5. **Sum Modulo \( p \):**
   - Since each pair sums to 0 modulo \( p \), the entire sum \( S \) is congruent to 0 modulo \( p \):
     \[
     S \equiv 0 \pmod{p}
     \]

6. **Prime Condition:**
   - Since \( S \) is a prime number and \( S \equiv 0 \pmod{p} \), \( S \) must be equal to \( p \) or a multiple of \( p \).
   - However, \( S \) is much larger than \( p \) for \( p > 2 \) because it includes terms like \( p^p \).

7. **Conclusion:**
   - Therefore, for \( p > 2 \), \( S \) cannot be a prime number because it is larger than \( p \) and divisible by \( p \).
   - Hence, the only prime number of the form \( 1 + 2^p + 3^p + \cdots + p^p \) is when \( p = 2 \).

The final answer is \( \boxed{ p = 2 } \) which gives \( 1 + 2^2 = 5 \).