The n-th prime is less than $n^2$? Let $p_n$ be the n-th prime number, e.g. $p_1=2,p_2=3,p_3=5$. How do I show that for all $n>1$, $p_n<n^2$?

You may use Bertrand's Postulate, which states that, for any natural number $n>1$, there exists a prime number $p$ such that $n<p<2n$; and induction. 
First note that the result does not hold for $p_1=2$. However it does hold that $p_2=3<2^2=4$. 
Now following an induction argument let's assume that for some fixed $n>1$ it holds that $p_n < 2^n$. Then by Bertrand's Postulate clearly $p_n < p_{n+1} < 2p_n$. So $p_{n+1}<2p_n < 2^{n+1}$. 
