I'm reading this (very short, 1 page long) paper by W.H. Mills where he determines that there exists a real number $A$ such that $f(n) = \lfloor {A^3}^n \rfloor$ is a prime number for all positive integers $n$.
In the paper before Theorem 1, he has this sequence $\{P_n\}_{n=0}^{\infty}$ with the property
$P_n^3 < P_{n+1} < (P_n + 1)^3 - 1. \hspace{20pt} (*)$
He defines a new sequence $\{u_n\}_{n=0}^{\infty}$ by $u_n = P_n^{3-n}$ and proceeds to prove things about that sequence. In particular he states (without proof) that $u_{n+1} > u_n$ (which can also be written P_{n+1}^{3-n-1} > P_n^{3-n}).
I am trying to justify the step and it is turning out more difficult than I thought it would be. I tried using $(*)$ in the following way:
$P_{n+1}^{3-n-1} > \left( P_n^3 \right)^{3-n-1} = P_n^{9-3n-3} = P_n^{6-3n}.$
To finish, I would need the inequality $P_n^{6-3n} > P_n^{3-n}$. I can see how to get $P_n^{6-3n} > P_n^{3-3n}$, but the sentence $P_n^{3-3n} > P_n^{3-n}$ is clearly false, so this will not work.
This indicates to me that I should start from the other side of $(*)$. If you try it using the fact that for $x>0$, $(x+1)^c > x^c+x^{c-1}+1$, this is what happens:
$\begin{array}{} (p_{n+1}+1)^{3-n} &> p_{n+1}^{3-n} + p_{n+1}^{3-n-1} + 1 \\ &> \left( p_n^{3-n} \right)^3 + \left( p_n^{3-n-1} \right)^3, \end{array}$
then I would like to conclude $> p_n^{3-n}$, but it is not going to be true since $p_n^{3-n} < 1$ implies $\left( p_n^{3-n} \right)^3 < p_n^{3-n} < 1$.
What am I not seeing?