Prove that the nth prime number $p_n$ (with $p_1 = 2, p_2 = 3, p_3 = 5$, etc.) satisfies $p_n \leq 2^{2n-1}$
So far I have figured out that $p_n$ = the nth prime and that I have to use mathematical induction to prove $p_n \leq 2^{2n-1}$. This is similar to the proof of infinitely many primes such that
$m = 1 + p_1 p_2 p_3\cdots p_n$ so there exists a prime $p | m$.
With this information I have concluded that $p_{n+1} \leq p \leq m = 1 + p_1 p_2 p_3\cdots p_n$
I need to figure out how to produce the right side of this inequality with induction. I'm not sure how to do this.
Thank you for any and all help.
Best Answer
A theorem(I forgot its name) states that there is always a prime between $n$ and $2n$.
Let’s jump to the inductive step.
Assume $$p_n \le 2^{2n-1}$$ is true.
Since there is a prime between $p_n$ and $2p_n$, $p_{n+1}<2p_n$.
So, $$p_{n+1}<2*2^{2n-1}<2^{2n}<2^{2(n+1)-1}$$
Therefore, $P(n+1)$ is true when $P(n)$ is true.
The base case is $n=1$, $p_1=2$.
By the principle of mathematical induction, the statement is true.