We have: $a_1=8p^m$, $a_2=3^{4p^m}$, $a_3=4^{3^{4p^m-1}}$, $\cdots$. So we guess that: $a_{n-1}|a_n-1,\forall n\geq 3$. Claim 1. $a_{n-1}|a_n-1,\forall n\geq 1$. Now, we will work with $p$ is prime number and $p|n$. If $p$ is odd, we have: $$v_p\left( a_n-1 \right)=v_p\left( (n+1)^{\frac{a_{n-1}}{n}}-1 \right)=v_p\left( n \right)+v_p\left( \dfrac{a_{n-1}}{n} \right)=v_p(n)+v_p(a_{n-1})-v_p(n)=v_P(a_{n-1})$$So $a_{n-1}|a_n-1$. If $p=2$, we also have: $$v_2(a_n-1)=v_2\left( (n+1)^{\frac{a_{n-1}}{n}}-1 \right)=v_2(n)+v_2(n+2)+v_2\left( \frac{a_{n-1}}{n} \right)-1=v_2(a_{n-1})+v_2(n+2)-1=v_2(a_{n-1})$$So that, $a_{n-1}|a_n-1,\forall n \geq 3$. Thus, $a_n\left(1-\frac{1}{a_1}\right)\left( 1-\frac{1}{a_2} \right)\cdots \left( 1-\frac{1}{a_n} \right) \in \mathbb{Z}$ if and only if $a_1|a_2-1 \Leftrightarrow 8p^m | 81^{p^m}-1$, so $p|81^{p^m}-1$. Claim 2. Given prime number $p$ and positive integer $m$. Then $81^{p^m}-81 \; \vdots \; p$. The first case: $m=1$, then $81^p-81 \; \vdots \; p$ (true as Fermat theorem). Suppose that $81^{p^m}-81 \; \vdots \; p$. We need to prove that: $81^{p^{m+1}}-81 \;\vdots \; p$. Easy to see that: $$81^{p^{m+1}}-81=81^{p^m\cdot p}-81=\left( 81^{p^m} \right)^p-81\equiv 81^p-91\equiv 0 \; [p].$$Return to solution, as $p|81^{p^m}-1$ and $p|81^{p^m}-81$ so $p|80 \Rightarrow p\in \left\lbrace 2;5 \right\rbrace$.