According to Bertrand's postulate, for every positive integer \(k\), there exists a prime \(p\) such that \(k \lt p \lt 2k\). Now let \(n = 2^{k-1}\). This means for every positive integer \(n \ge 2\), there exists a prime \(p\) such that \(2^{k-1} \lt p \lt 2^k\).
Another way of saying this is that in every interval of the form \((2^{k-1}, 2^k)\) for \(k=2,3,4,\ldots\), there exists at least one prime.
In the interval \((2^1,2^2)\), there is the prime 3. In the interval \((2^2,2^3)\), there is the prime 5. In the interval \((2^3,2^4)\), there is the prime 11.
If the intervals \((2^1,2^2)\) and \((2^2,2^3)\) contain at least one prime in each, then the interval \((2^1,2^3)\) contains at least two primes. Similarly, \((2^1,2^4)\) contains at least three primes and \((2^1,2^5)\) contains at least four primes. In general, \((2,2^k)\) contains at least \(k-1\) primes.
Since 2 it self is a prime, then the interval \((1,2^k)\) contains at least \(k\) primes. Another way of saying this is that there are at least \(k\) primes less than \(2^k\), where \(k \ge 2\). This also means that the \(k\)th prime is \(\lt 2^k\), where \(k \ge 2\).
What about a case where \(k=1\)? We are considering only one prime (2), and this is equal to \(2^1\). This means that the \(k\)th prime is \(\le 2^k\), where \(k \ge 1\).