How to show that if $P_n^{3-n}< x<(P_{n}+1)^{3-n}$ then $P_n I was reading Mills' paper where he gives the original derivation of the prime-representing function named after him. In the proof he defines a sequence of primes numbers  $\left\{P_n\right\}_{n \in \mathbb{N}}$ such that
$$
P_n^3 < P_{n+1} < (P_{n}+1)^3 -1, \ \  \forall n \in \mathbb{N}
$$
After this he claims the following:

*

*$$P_n^{3-n}<P_{n+1}^{3-n-1} $$

*$$ (P_{n+1}+1)^{3-n-1}<(P_n+1)^{3-n}$$

*If $P_n^{3-n}< x<(P_{n}+1)^{3-n}$ then $$P_n <x^{3^n} <P_{n}+1$$
and I attempted to prove them.

For the first one I tried $$P_n^3 <P_{n+1} \color{darkblue}{\implies} P_n^{3-n} <P_{n+1}^{\frac{3-n}{3}} $$ and if we set $\frac{3-n}{3}< 3-n-1$ this has solution $n < \frac32$, which doesn't help. For the second one I found basically the same issue. For the third one it's easy to get that $P_n < x^{\frac{1}{3-n}} < P_n +1$, but I can't see a wat to get a $x^{3^n}$ without just elevating the entire inequality to the $3^n$ power, which doesn't help to isolate $P_n$ or $P_n +1$.

How can I show that these implications hold? Or if I misunderstood what the author was saying (which is also likely), what am I doing wrong?

Thank you!
 A: So after trying a bunch of different things, I'm pretty sure that all the $3-n$ exponents are actually meant to be $3^{-n}$, since if it's interpreted as the latter all the properties do follow pretty directly.
The confusion arises because in some parts of the paper an expression like $a^{b^c}$ has a clear size and position difference between the $3$ levels of exponentiation, but this isn't written in the same way in the inequalities I wrote in the question. Here's a side-by-side of what I mean:

On the right, $3^n$ does look like exponentiation as the $n$ is higher than $3$, but on the left the $-n$ looks to be the same size and on the same height as the $3$.

Assuming that the expressions are meant to be $3^{-n}$ instead of $3-n$, then the questions I ask become the following: If $P_n^3 < P_{n+1} < (P_{n}+1)^3 -1$ then

*

*Show $P_n^{3^{-n}}<P_{n+1}^{3^{-n-1}}$.

Proof: Since
$$
\color{purple}{P_n^3} < \color{purple}{P_{n+1}} \color{darkblue}{\implies}P_n^{3^{-n}} = \left(\color{purple}{P_n^3}\right)^{3^{-n-1}} < \left(\color{purple}{P_{n+1}}\right)^{3^{-n-1}}  = P_{n+1}^{3^{-n-1}}
$$$□$

*

*Show $(P_n+1)^{3^{-n}}>(P_{n+1}+1)^{3^{-n-1}}$.

Proof: Since
\begin{align*}
  (P_{n}+1)^3 -1> P_{n+1} & \color{darkblue}{\implies} \color{purple}{(P_{n}+1)^3} > \color{purple}{ P_{n+1}+1}\\
 & \color{darkblue}{\implies}(P_{n+1}+1)^{3^{-n}} = \left(  \color{purple}{(P_{n}+1)^3} \right)^{3^{-n-1}} > \left( \color{purple}{P_{n+1}+1}\right)^{3^{-n-1}}  = (P_n+1)^{3^{-n-1}}
\end{align*}$□$

*

*If $P_n^{3^{-n}}<x <\left(P_n+1\right)^{3^{-n}}\color{darkblue}{\implies} P_n <x^{3^n} <P_{n}+1 $.

Proof: Taking the $3^n$'th power of the hypothesis gives the result.
$□$
And since all these proofs are so direct and (I believe) don't clash with any other part of the argument in Mills' proof, I think this is the intended argument.
