Problem 1541

1. Assume the contrary: Suppose there exists some $n \in \mathbb{N}_+$ such that all prime factors of $a_n$ are also prime factors of $\prod_{k=1}^{n-1} a_k$.

2. Choose a prime factor: Let $p$ be a prime factor of $a_n$. Then there exists $1 \leq j \leq n-1$ such that $p \mid a_j$. Choose the smallest such $j$.

3. **Prove periodicity modulo $p$**: We need to show that $j \mid n$. Consider the sequence modulo $p$:
$$a_{j+1} \equiv a_j^d + c \equiv c \equiv a_1 \pmod{p}$$
By induction, we can show that for any $u \in \mathbb{N}_+$,
$$a_{j+u} \equiv a_u \pmod{p}$$
This implies that the sequence $\{a_n\}$ is periodic with period $j$ modulo $p$.

4. **Express $n$ in terms of $j$**: Write $n = qj + r$ where $q \in \mathbb{Z}$ and $0 < r \leq j$. Then,
$$a_n \equiv a_r \pmod{p}$$
By the minimality of $j$, we must have $r = j$, hence $j \mid n$.

5. **Prove $v_p(a_n) = v_p(a_j)$**: Let $v_p(a_n) = r$. Then,
$$a_{j+1} = a_j^d + c \equiv c \equiv a_1 \pmod{p^{rd}}$$
By induction, for any $u \in \mathbb{N}_+$,
$$a_{j+u} \equiv a_u \pmod{p^{rd}}$$
Thus,
$$a_n \equiv a_j \pmod{p^{rd}}$$
Since $v_p(a_j) = r < rd$, it follows that $v_p(a_n) = v_p(a_j)$.

6. Contradiction: For all prime factors $p$ of $a_n$,
$$v_p(a_n) \leq v_p\left(\prod_{k=1}^{n-1} a_k \right)$$
Hence,
$$a_n \mid \prod_{k=1}^{n-1} a_k$$
Therefore,
$$a_n \leq \prod_{k=1}^{n-1} a_k$$
However, for $n = 2$,
$$a_2 = a_1^d + c > a_1$$
By induction, if $a_{n-1} > \prod_{k=1}^{n-2} a_k$, then
$$a_n > a_{n-1}^2 > \prod_{k=1}^{n-1} a_k$$
This shows that for all $n \in \mathbb{N}_+$,
$$a_n > \prod_{k=1}^{n-1} a_k$$
This is a contradiction. Therefore, there exists a prime factor $p$ of $a_n$ such that $p \nmid \prod_{k=1}^{n-1} a_k$.

$\blacksquare$