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 $n&gt;1$, $p_n&lt;n^2$?
In Zagier's the first 50 million prime numbers a very elementary proof is given that for $n &gt; 200$ we have

$$\pi(n) \ge \frac23 \frac{n}{\log n}$$

where $\pi(x)$ is the number of primes below $x$ (as well as a bound in the other direction). In fact, it already holds for $n \ge 3$, as can be directly checked. 

Suppose that $p_n &gt; n^2$, then for this $n$ we have $\pi(n^2) &lt; n$, but that violates the bound already for $n = 5$.
You may use Bertrand's Postulate, which states that, for any natural number $n&gt;1$, there exists a prime number $p$ such that $n&lt;p&lt;2n$; and induction. 

First note that the result does not hold for $p_1=2$. However it does hold that $p_2=3&lt;2^2=4$. 

Now following an induction argument let's assume that for some fixed $n&gt;1$ it holds that $p_n &lt; 2^n$. Then by Bertrand's Postulate clearly $p_n &lt; p_{n+1} &lt; 2p_n$. So $p_{n+1}&lt;2p_n &lt; 2^{n+1}$.