Let $a_{1}=2$, $a_{n+1}=2^{a_{n}}$ for $n \geq 1$
Let $b_{1}=3$, $b_{n+1}=3^{b_{n}}$ for $n \geq 1$
Is is true that $a_{n+2}>b_{n}$ for all $n \geq 1$?
If so, is the proof elementary? (Use only Mathematical techniques covered in an undergraduate degree)
Yes. Here is a simple proof by induction:
Theorem. $a_{n+2} > 4 b_n$
If $n=1$, then $a_3 = 16 > 4*3 = 4b_1$.
Now assume $a_{n+2} > 4 b_n$:
$$a_{n+3} =2^{a_{n+2}} > 2^{4b_n} = 16^{b_n} > 4^{b_n} * 3^{b_n} > 4 * 3^{b_n} = 4b_{n+1}$$
and the result follows.