I've been checking my university's past exams and I've found the following problem:
Let $p_n$ be the $n-$th prime. Prove that: $p_n\le2^{2^{n-1}}$.
The main idea I'm trying to prove is that $p_n^2\ge p_{n+1}$, I know that Bertrand's postulate give the much better bound $2p_n\ge p_{n+1}$ but I'm trying to stay in the course's context.
Of course this translates to the existence of a prime in $\{p_n+1, \cdots, p_n^2\}$. We'll show that for each positive integer $n$ there exists a prime in $\{n+1, \cdots, n^2\}$.
Mimicking Erdös's proof of Bertrand's postulate; We consider the binomial coefficient $\binom{n^2}{n}$, each prime that divides it satisfies: $\displaystyle p^{v_p\binom{n^2}{n}}\le n^2$, by Legender's formula. Supposing that no prime is in: $\{n+1, \cdots, n^2\}$, and using the fact that there are at most $\displaystyle \frac{n+1}2$ primes in $\{1, \cdots, n\}$, we get: $\binom{n^2}{n}\le (n^2)^{\frac{n+1}2}=n^{n+1}$, which fails for $n\ge 3$.
My question is: is there an easier method to prove that $p_n^2\ge p_{n+1}$?
