# Inequality with prime numbers

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?

• I think this follows from Bertrand's Postulate...are you allowed to assume that? Sep 6, 2014 at 23:08
• Bertrand's postulate is proofed, so We can use it :) But the best is the easiest proof. Sep 6, 2014 at 23:17
• There are many results about the bounds of the $n$th prime. They perhaps are helpful, chee up :)
– Yes
Sep 6, 2014 at 23:30
• @TheKwiatek666 So you want a proof that does not use Bertrand's Postulate? Sep 7, 2014 at 0:08

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 $k \in \mathbb{N}$.