Let $m$ be a positive integer and $a$ a positive real number. Find the greatest value of $a^{m_1}+a^{m_2}+\ldots+a^{m_p}$ where $m_1+m_2+\ldots+m_p=m, m_i\in\mathbb{N},i=1,2,\ldots,p;$ $1\leq p\leq m, p\in\mathbb{N}$. 
1. **Understanding the Problem:**
   We need to find the maximum value of the sum \(a^{m_1} + a^{m_2} + \ldots + a^{m_p}\) where \(m_1 + m_2 + \ldots + m_p = m\) and \(m_i \in \mathbb{N}\) for \(i = 1, 2, \ldots, p\), with \(1 \leq p \leq m\).

2. **Key Insight:**
   The key insight is to use the inequality \(a^{x+y-1} + a \ge a^x + a^y\) for \(x, y \ge 1\). This can be shown as follows:
   \[
   a^{x+y-1} + a \ge a^x + a^y \Leftrightarrow a(a^{x-1} - 1)(a^{y-1} - 1) \ge 0
   \]
   Since \(a \ge 1\), \(a^{x-1} \ge 1\) and \(a^{y-1} \ge 1\), the product \((a^{x-1} - 1)(a^{y-1} - 1) \ge 0\) is non-negative.

3. **Applying the Insight:**
   We can use this inequality to combine terms in the sum \(a^{m_1} + a^{m_2} + \ldots + a^{m_p}\). By repeatedly applying the transformation \((m_1, m_2) \rightarrow (m_1 + m_2 - 1, 1)\), we can reduce the number of terms in the sum while ensuring the sum does not decrease.

4. **Optimal Configuration:**
   By continuing this process, we find that the optimal configuration is to have \(p-1\) terms equal to 1 and one term equal to \(m - (p-1)\). This is because:
   \[
   a^{m_1} + a^{m_2} + \ldots + a^{m_p} \leq (p-1)a + a^{m - (p-1)}
   \]
   This configuration maximizes the sum.

5. **Conclusion:**
   The greatest value of \(a^{m_1} + a^{m_2} + \ldots + a^{m_p}\) is given by:
   \[
   (p-1)a + a^{m - (p-1)}
   \]

The final answer is \( \boxed{ (p-1)a + a^{m - (p-1)} } \).