I'm looking to prove that:
$$p_n \leq 2^{2^{n-1}}$$
Where $p_n$ denotes the $n$th prime in ascending order. The proof method is induction. I've solved my base case, that is: $n=1$ $p_1 = 2$, $2^{2^0}=2$, $2\leq2$ Therefore $P(1)$ is true.
And the induction hypothesis is that $P(1), P(2),\ldots P(k)$ is true for some integer $k$.
I'm stuck trying to prove $P(k+1)$, or $p_{k+1} \leq 2^{2^k}$
The closest I have come was rewriting $2^{2^k}$ as $(2^{2^{k-1}})^2$, which is the square of the value of $P(k)$. My guess at the moment is that if I can prove that a prime is strictly less then the square of the previous prime then my proof would be complete, but I can't seem to do so.
Thanks in advance for help.