Let $ S_k = 1^{a_k} + 2^{a_k} + \ldots + p^{a_k}$. Observe that $ f(a_1, a_2, \ldots, a_m)$ factors as $ S_1 S_2 \ldots S_m$. Thus, $ p | f(a_1, \ldots, a_m)$ if and only if $ p$ divides one of the terms in the sum. Clearly, if some $ a_k$ is 0, then $ p | S_k$. Thus, we may suppose each of $ a_1, a_2, \ldots, a_m$ are nonzero. Lemma: If $ a_k > 0$, and $ p - 1 | a_k$, then $ p | S_k + 1$. Otherwise, if $ a_k > 0$ then $ p | S_k$ Proof: If $ p - 1 | a_k$, the result follows easily from Fermat's little theorem. Otherwise, let $ g$ be a generator mod $ p$. Observe that $ S_k \equiv \sum{i=1}^{p-1} i^k \equiv \sum_{i=0}^{p-2} (g^i)^k \equiv \frac{(g^k)^{p-1} - 1}{g^k - 1} \pmod p$. Since $ g^k \not \equiv 1 \pmod p$ (since $ p - 1 \not | k$), and by Fermat's little theorem $ (g^k)^{p-1} - 1$ is a multiple of $ p$, $ S_k$ is a multiple of $ p$, as desired. With the lemma, it follows that $ p | f(a_1, \ldots, a_m)$ if and only if some one of $ a_1, a_2, \ldots, a_m$ is 0 or not a multiple of $ p - 1$. It is unfortunate that my final answer was incorrect for this problem... oh well.