I found exercise in my book for number theory that I can't resolve. How do you show that $$p_n < e^{1+n}$$ where $p_n$ is $n$-th prime number?
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$\begingroup$ I think this follows from Bertrand's Postulate...are you allowed to assume that? $\endgroup$– NishantSep 6, 2014 at 23:08
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$\begingroup$ Bertrand's postulate is proofed, so We can use it :) But the best is the easiest proof. $\endgroup$– TheKwiatek666Sep 6, 2014 at 23:17
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$\begingroup$ There are many results about the bounds of the $n$th prime. They perhaps are helpful, chee up :) $\endgroup$– YesSep 6, 2014 at 23:30
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$\begingroup$ @TheKwiatek666 So you want a proof that does not use Bertrand's Postulate? $\endgroup$– graydadSep 7, 2014 at 0:08
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Let's try induction. For the base case of $n=1$, it is clear that $p_{1}=2<e^{1+1}=e^2$. Now let's suppose for all $k$ where $n \geq k >1$ that our result holds. We know by Bertrand's Postulate that $p_{k+1}<2p_{k}$ for all $k \in \mathbb{N}$ and by induction we know $2p_{k}<2e^{k+1}$. It follows that $$p_{k+1}<2e^{k+1}<(e)e^{k+1}=e^{k+2}$$ We now know our result holds for all $k \in \mathbb{N}$.
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$\begingroup$ This is one of the first exercises in this book. Maybe is it possible to resolve that without Bertrand's Postulate? This theorem is proofed four chapters later. $\endgroup$ Sep 7, 2014 at 14:17