Bounds on the $n$-th prime

Induction base. $p_1=2\le2$.

Inductive step. $p_{n+1}$ divides $p_1\cdots p_n+1$ (or is less than it), so $p_{n+1}\le p_1\cdots p_n+1$. By induction, $p_i\le2^{2^{i-1}}$, so
$$p_{n+1}\le\prod_{i=1}^n2^{2^{i-1}}+1=2^{2^n-1}+1<2^{2^n}.$$

Conclusion. $p_n\le2^{2^{n-1}}$.

Remark. Much better bounds exist: $p_n\sim n\ln n$ by the prime number theorem.