Let $x_0 = 0$ and $x_n = a_n/c$. It is easy to see that $x_1 = 1$, $x_2 = 2c^2 + 1$, and
$$x_{n+1} = c^2(x_n^3 - 4x_n^2 + 5x_n) + 1. \quad (1)$$
It is clear that $x_n$ is an increasing sequence. To prove the original statement, it suffices to prove the corresponding statement for the sequence $x_n$.

We first prove three lemmas:

(1) Lemma 1: If $i = j \pmod m$, then $x_i = x_j \pmod{x_m}$.
This lemma is equivalent to $x_{i+m} = x_i \pmod{x_m}$. Fix $m$; this is clearly true for $i = 0$. If $x_{i+m} = x_i \pmod{x_m}$, then we have
$$\begin{aligned} x_{i+m+1} &= c^2(x_{i+m}^3 - 4x_{i+m}^2 + 5x_{i+m}) + 1 = c^2(x_i^3 - 4x_i^2 + 5x_i) + 1 \\ &= x_{i+1} \pmod{x_m} \end{aligned}$$
Hence by mathematical induction, the proof is complete.

(2) Lemma 2: If $i, j \ge 2$ and $i = j \pmod m$, then $x_i = x_j \pmod{x_m^2}$.
This lemma is equivalent to $x_{i+m} = x_i \pmod{x_m^2}$. Fix $m$, and note that this holds for $i = 2$. Proceeding by induction in a manner similar to the previous item yields the result.

(3) Lemma 3: For all $n \ge 2$, we have $x_n > x_1x_2\cdots x_{n-2}$.
Note that the lemma is clearly true for $n=2,3$. For $n>3$, by monotonicity we know $x_n > 7$, so
$$x_{n+1} > x_n^3 - 4x_n^2 + 5x_n > 7x_n^2 - 4x_n^2 > x_n^2 > x_n x_{n-1},$$
hence by mathematical induction, the proof is complete.

Returning to the original problem. By Lemma 3, there exists a prime $p$ and a positive integer $t$ such that $p^t$ divides $x_n$ but does not divide $x_1x_2\cdots x_{n-2}$. We will now show that this $p$ is the $p$ required by the problem.
If not, let $k$ be the smallest positive integer satisfying $p|x_k$. By Eq. (1), $x_{n-1}$ and $x_n$ are coprime, and $x_1 = 1$, so we have $2 \le k \le n-2$. Write $n = qk+r$, where $q \ge 0$ and $0 \le r < k$. By Lemma 1, we know that $x_n$ and $x_r$ are congruent modulo $x_k$, so $p|x_r$; but by the minimality of $k$, this means $r=0$, and hence $k|n$.
Now, by Lemma 2, we have that $x_n$ and $x_k$ are congruent modulo $x_k^2$. Let $\alpha \ge 1$ be the maximum value such that $p^\alpha|x_k$. By the argument above, we know $p^{\alpha+1}|x_n$, while $p^{2\alpha}|x_k^2$. But this forces $p^{\alpha+1}|x_k$, contradicting the maximality of $\alpha$. Contradiction! This completes the proof.