Prove sequence using induction

$a_1=1$, $a_{n+1} = 3 a_n^2$.

Prove for all positive integers, $a_n\leq{3^{2^n}}$ using induction.

My work so far:

Base case is true (1 < 9)

Induction Hypothesis: $a_k\leq{3^{2^k}}$

IS: prove that n = k+1 is true

I'm stuck because I just can't seem to prove the induction step. Any help is appreciated.

Answer

We need to show the stronger condition

$$a_n\leq{3^{2^n-1}}(\leq{3^{2^n}})$$

and therefore assuming as Induction Hypothesis $a_k\leq{3^{2^k-1}}$ we have

$$a_{k+1}=3a_k^2\stackrel{Ind. Hyp.}\leq 3\cdot (3^{2^{k}-1})^2={3^{2^{k+1}-1}}$$

Refer also to the related