How might I prove or disprove this conjecture: $7p_{k-2} < 8p_{k-4}$ for all $k \ge 35$? Patterns observed while working with recursive prime factorizations (arXiv:2102.02777) lead me to the following conjecture (which is a specific case of a much more general conjecture that I am not presently prepared to state):
Let $k$ be an integer greater than or equal to 33.  Then $7p_{k} < 8p_{k-2}$, where $p_i$ is the $i$th prime number.
@Lulu and @Peter have been very helpful in pointing me to a solution for that conjecture.  Having experienced so much success, I venture to ask about another special case of the generalized conjecture (please forgive me if I do not immediately see that the answers already given are sufficient to address it):
Let $k$ be an integer greater than or equal to 50.  Then
$p_k < 2p_{k-1} < 3p_{k-1} < 4p_{k-2} < 5p_{k-1} < 6p_{k-2} < 7p_{k-1} < 8p_{k-3} < 9p_{k-2} < 10p_{k-2}$.
Since @lulu was kind enough to subsequently point out that Bertrand's Postulate is sufficient to prove that latter conjecture, I am now ready to present the generalization of the conjecture (note that I begin with $p_{k-1}$ rather than $p_{k}$ to show successive terms in OEIS sequence A073093 as the subtrahend in the subscript):
Let $n \in \mathbb{N}_{+}$.  Then there exists a positive integer $k_{\text{min}}$ such that for all integers $k \ge k_{\text{min}}$,
$p_{k-1} < \cdots < np_{k-a_{n}},$
where $a_{n}$ is the number of prime power divisors of $n$, i.e., $a$ is Sequence A073093 in the Online Encyclopedia of Integer Sequences (OEIS).
By the way, I cannot figure out what the pattern is in the sequence of $k_{\text{min}}$ for successive values of $n$; that sequence is $(2,3,8,8,13,13,35,35,10,14,\ldots)$.  Can anyone identify a formula for the sequence?
 A: EDIT: As has been pointed out in the comments, there is an error in this proof (basically from copying the bounds incorrectly from wikipedia) which means the details of the argument are wrong. However, the overall gist of the argument is correct so I think it is worth it to leave this answer up. Just note that the bounds should be
$$k\ln(k)+k\ln(\ln(k))-k<p_k<k\ln(k)+k\ln(\ln(k))$$
where I had accidentally copied
$$k\ln(k)+k\ln(\ln(k))-1<p_k<k\ln(k)+k\ln(\ln(k))$$
With this fix, the $K$ described below should be $611$ instead of the $27$ that I originally got.

As has been mentioned in the comments, this conjecture seems true but case checking might be unavoidable. To prove its true for some $k\in\mathbb{N}$, I use the following bounds
$$k\ln(k)+k\ln(\ln(k))-1<p_k<k\ln(k)+k\ln(\ln(k))$$
for $k\geq 6$. Then
$$\frac{7p_k}{8p_{k-2}}<\frac{7}{8}\cdot \frac{k\ln(k)+k\ln(\ln(k))}{(k-2)\ln(k-2)+(k-2)\ln(\ln(k-2))-1}$$
We must now find some $K\in\mathbb{N}$ such that $k\geq K$ implies
$$\frac{7}{8}\cdot \frac{k\ln(k)+k\ln(\ln(k))}{(k-2)\ln(k-2)+(k-2)\ln(\ln(k-2))-1}<1$$
Of course, such a $K$ must exist since
$$\lim_{k\to\infty}\frac{7}{8}\cdot \frac{k\ln(k)+k\ln(\ln(k))}{(k-2)\ln(k-2)+(k-2)\ln(\ln(k-2))-1}=\frac{7}{8}$$
the only difficulty being finding such a $K$. Okay, here is a method for finding such a $K$. Define
$$f(x)= \frac{x\ln(x)+x\ln(\ln(x))}{(x-2)\ln(x-2)+(x-2)\ln(\ln(x-2))-1}$$
Then differentiating gives two awful functions in the numerator and denominator
$$f'(x)=\frac{g(x)}{h(x)^2}$$
(I haven't written them here since they are rather large and unweildy and don't really provide any insight into the problem). Since $h(x)^2>0$, it is sufficient to find $X$ such that $x\geq X$ implies $g(x)<0$. Again, after lots of simplification, we find that
$$g(x)<-2\ln(\ln(x))^2$$
(with the minor stipulation that $x\geq 3^3$ in order for some of the inequalities we used to work). That is, the derivative of $f(x)$ is always negative for $x\geq 3^3$. This basically completes the proof since
$$\frac{7}{8}f(27)=0.975<1$$
Thus, a sufficient bound is $K=27$ which is already less than what you need. This checks well with the actual bound, since the final prime the conjecture is not true for is $k=25$. That is
$$\frac{7p_{25}}{8p_{23}}=\frac{679}{664}>1$$
and the actual bound is $K=26$.

FINAL EDIT: This edit will prove the existance of $K(n)$ such that
$$p_k<2p_{k-1}<3p_{k-2}<...<np_{k-a_n}$$
where $a_n$ is the number of non-distinct prime factors in $n$, is true for all $k\geq K(n)$
By the prime number theorem, we have that
$$\lim_{k\to\infty}\frac{ap_{k}}{bp_{k-r}}=\lim_{k\to\infty}\frac{ak\ln(k)}{b(k-r)\ln(k-r)}=\frac{a}{b}$$
(note that $r\in\mathbb{Z}$). Thus, $K(2)$ exists since
$$\lim_{k\to\infty}\frac{p_k}{2p_{k-1}}=\frac{1}{2}$$
(in fact, $K(2)=2$ follows from Bertrand's postulate). Now, to find $K(n+1)$ from $K(n)$, define $K(n+1)$ as the maximum of $K(n)$ and $K^{'}$. Here, $K^{'}$ is defined as the smallest natural such that $k\geq K^{'}$ implies
$$\frac{np_{k-a_n}}{(n+1)p_{k-{a_{n+1}}}}<1$$
Again, we know such a $K^{'}$ must exist from the prime number theorem. In this manner, all $K(n)$ can be defined. Of course, finding these $K(n)$ and proving that they are certain values is a much more difficult challenge.
