Let $p_n$ denote the $n^{th}$ prime then prove that $p_n<2^n$, where $n≥2$

Question

let us prove it by induction if n=2 then p2p2 =3 which is less than 2^2 if n=3 then p3p3 =5 which is less than 2^3 similarly it is true for p= n now if I take n= n+1, then pn+1pn+1 is less than 2^(n+1)= 2^nx2 so in this case this is also valid So tell me is this correct

Answer 1

Let us prove it by induction:

If $n=2$ then $p_2 = 3 \lt 2^2$.
If $n=3$ then $p_3 = 5 \lt 2^3$.
Assume it is true for $n=k$, while $k\ge 2$. Then $p_k\lt 2^k$.
By Bertrand's postulate , there is a prime $p$ such that $2^k\lt p\lt 2^{k+1}$, so $p_{k+1}\lt 2^{k+1}$

The theorem follows.