A question on an inequality appearing in the proof of Mills' theorem I've been reading Mills' proof that there exists a real number $A$ such that $\lfloor A^{3^n}\rfloor$ is a prime number for all integers $n\ge 1$.
There is a sentence that goes:

Then by the lemma we can construct an infinite sequence of primes, $P_0,P_1,P_2,\;\cdots$, such that ${P_n}^3<P_{n+1}<(P_n+1)^3-1$.

I understand that from the lemma it obvioulsy follows that for every sufficiently large prime number $P$ there is a prime number $q$ such that $P^3<q<(P+1)^3-1$, but I don't understand why ${P_n}^3<P_{n+1}$.
Thanks in advance!!
 A: Let's just start by stating the lemma again :

For every sufficiently large prime number $P$ there is a prime number $q=f(P)$ such that $P^3<q<(P+1)^3-1$. the prime $q$ depends in the $q$ we have chosen.

Now what the sentence affirms is that we can construct an infinite sequence of primes $p_0,...,p_n$ such that $p_n^3\leq p_{n+1}\leq (p_n+1)^3-1$. This follows directly from the lemma  as you can always apply the lemma for $p_n$ in order to construct $p_{n+1}$ so $p_{n+1}=f(p_n)$, obviously $p_{n+1}$ is not the (n+1)-th prime but it's the prime $q$ which verifies $p_n^3<q<(p_n+1)^3-1$ which always exists according to the lemma.
The complete proof goes is as follow:


*

*First take $p_0$ a large prime (than $K^3$), using the lemma for $p_0$ there exists $q$ a prime such that $p_0^3\leq q\leq (p_0+1)^3-1 $, let's take $p_1=q$, so we have constructed $p_0,p_1$ such that $p_0^3<p_1<(p_0+1)^3-1$, if you do the same for $p_1$ you will construct $p_2$ and for $p_2$ you construct $p_3$ $etc\cdots$, that's what the second part explains.

*Now Assume that you constructed $p_0,p_1,....,p_n$ ($n\geq 2$)such that for every $0\leq i\leq n-1$ we have $p_i^3\leq p_{i+1}\leq (p_i+1)^3-1$, let $p_{n+1}$ be the prime $q$ such that $p_n^3\leq q\leq (p_n+1)^3-1$ which exists according to the lemma. hence we have constructed $p_0,p_1,....,p_{n+1}$ such that for every $0\leq i\leq n$ we have $p_i^3\leq p_{i+1}\leq (p_i+1)^3-1$


This argument proofs that we can always add another element to the sequence $p_0,...,p_n$ this means that can extend this sequence to an infinite sequence of prime verifying the property.
I hope it's clear now.
