Define $b_n = (a_1 + 1)(2a_2 + 1) \cdots ((n-1)a_{n-1} + 1)$ for $n = 2, 3, \dots$ and let $b_1 = 1$. By using mathematical induction we will prove that for all positive values of $n$
$$b_{n+1} = (b_1 + 2b_2 + \dots + n b_n)^2 - 5 \quad (1)$$
For $n=1$ we have $b_2 = a_1 + 1 = -4$ and hence $b_2 = -4 = b_1^2 - 5$ and (1) holds. Assume that (1) holds for $n = k - 1$. Then we have $b_k = (b_1 + 2b_2 + \dots + (k-1)b_{k-1})^2 - 5$. In order to prove that (1) is held for $n = k$ we show that
$$b_{k+1} - b_k = k^2 b_k^2 + 2k b_k (b_1 + 2b_2 + \dots + (k-1)b_{k-1}) \quad (2)$$
By definition of $b_n$, we have $a_{n+1} - a_n = (a_1 + 1)(2a_2 + 1)(3a_3 + 1) \cdots ((n-1)a_{n-1} + 1)((n^2 + n)a_n + 2n + 1) = (n+1)b_{n+1} + n b_n$. Hence we obtain that
$$a_k - a_2 = \sum_{i=3}^{k} a_i - a_{i-1} = \sum_{j=3}^{k} j b_j + (j-1) b_{j-1} = k b_k + 2(3b_3 + 4b_4 + \dots + (k-1)b_{k-1}) + 2b_2.$$
Since $a_2 = -6$, $b_1 = 1$, $b_2 = -4$, we conclude that $a_k - k b_k = 2(b_1 + 2b_2 + \dots + (k-1)b_{k-1})$. Therefore
$$\frac{b_{k+1} - b_k}{k b_k} - k b_k = 2(b_1 + 2b_2 + \dots + (k-1)b_{k-1})$$
and (2) is held. Thus, (1) is proved. Now if $p$ is a prime number dividing $n a_n + 1$, then $p$ also divides $b_{n+1}$ and therefore we can choose $m = b_1 + 2b_2 + \dots + n b_n$ and in this case $p$ divides $m^2 - 5 = b_{n+1}$.