Let $ n_1=1,n_{i+1}=a^{n_i}+1$ is sequence of odd numbers We have $ n_1|a^{n_1}+1$ and if $n_i|a^{n_i}+1=n_{i+1}$ then $\frac{a^{n_{i+1}}+1}{n_{i+1}}=\frac{a^{n_{i+1}}+1}{a^{n_i}+1}$ is integer because $n_i|n_{i+1}$

Since $a$ is even, $a+1$ will always have odd prime factor $p$. Then by LTE, we have $$v_p(a^{p^k}+1)=v_p(a+1)+k>k\implies p^k\mid a^{p^k}+1$$for all positive integers $k$, and we're done.