a sequence question

Question

A sequence is defined such that $a_1=1$, $a_2=2$, and $a_{n+2}=a_{n+1}^{a_n}$ for $n\ge 1$. What is the 9th term of this sequence?

(A) $2^{16}$ (B) $2^{256}$ (C) $2^{280}$ (D) $2^{2^{256}}$ (E) $2^{2^{280}}$

After noting that there number of 2s of the two previous term adds up to the next term, I find out the solution should be $2^{1024*1024}$ but apparently its is way too big. Any hint will be appreciated.

Answer 1

Let's check term by term,

$a_3={a_2}^{a_1}=2^1=2$.

$a_4={a_3}^{a_2}=2^2$.

$a_5={a_4}^{a_3}=({2^2})^2={2^{2^2}}=2^4$.

$a_6={a_5}^{a_4}=({{2^2}^2})^{2^2}={2^{2^{4}}}=2^{16}$.

$a_7={a_6}^{a_5}=({2^2}^4)^{2^4}={2^{2^{8}}}=2^{256}$.

$a_8={a_7}^{a_6}=({2^2}^8)^{2^{16}}={2^{{2}^{24}}}$.

$a_9={a_8}^{a_7}=({{2^2}^{24}})^{2^{256}}={2^{2^{280}}}$.

Hence correct answer is (E).