Bump .

Recall that $a^{pq}-b^{pq}=(a^p-b^p)(\dots)$. We proceed by induction. Base case: We need to prove $a^{a^a}-a^a|a^a-a$. That's true because $a^{a^a}-a^a=a^a(a^{a^a-a}-1), a^a-a=a(a^{a-1}-1)$, and $a-1|a^a-a, a|a^a$ Induction step: we know $(a_n-a_{n-1})|(a_{n+1}-a_n)$, so $a^{a_n-a_{n-1}}-1|a^{a_{n+1}-a_n}-1$. Hence $\frac{a_{n+1}}{a_n}-1|\frac{a_{n+2}}{a_{n+1}}-1$. Also obviously $a_n|a_{n+1}$, so we can multiply first number by $a_n$ and second by $a_{n+1}$, so done.