Is this a valid partial refinement of Ingham's upper bound for prime gaps?

Question

This is a follow-up question from this one that was kindly answered by @JordanPayette. The corrections were applied for this solution.

Let $p_n$ denote the $n^{th}$ prime number. Ingham showed that:

$$p_{n+1} - p_n \lt K p_n^{\frac{5}{8}}$$

where $K$ is a fixed positive integer, is an upper bound for the prime gaps.

(A.E.Ingham, On the difference between consecutive primes, Quart. J. Math. Oxford Ser. vol. 8 (1937) pp. 255-266)

I would like to know if the following easy manipulations would be valid to reduce the value $\frac{5}{8}$ up to $\frac{1}{8}$. The steps on $\color{blue}{blue\ color}$ have been verified in the previous question, so if possible I would appreciate very much if somebody could kindly verify if the rest are also accurate (the questions are at the end).

1. $\color{blue}{(verified)}$ Both sides to the eighth:

$$(p_{n+1} - p_n)^8 \lt K^8 p_n^{5}$$

2. $\color{blue}{(verified)}$ Def. $K_2 = K^8$ a new fixed constant $K_2 \gt K$:

$$(p_{n+1} - p_n)^8 \lt K_2 p_n^{5}$$

3. $\color{blue}{(verified)}$ By Fermat's Little Theorem, we know that $p_n^{5}$ can be replaced as follows}:

$$p_n^{5}= 5K^{'} + p_n$$

For a given positive unbounded constant $K^{'}$. Thus:

$$(p_{n+1} - p_n)^8 \lt K_2 (5K^{'} + p_n)$$

But (credits for the following explanation to @JordanPayette) the map

$$ K' : \mathbb{P} \to \mathbb{N} : p \mapsto \frac{p^5 - p}{5}$$

is not constant and unbounded from above.

---

$\color{red}{From\ this\ point\ ahead\ the\ manipulations\ need\ to\ be\ verified}$.

The right side of the inequality can be replaced as follows:

$$(p_{n+1} - p_n)^8 \lt K_2 (5K^{'} + p_n) = (K_2 \cdot 5K^{'}) + K_2 p_n$$

Our partial refinement will be true only if the constants are fixed and bounded, so we need to add a condition to handle $(K_2 \cdot 5K^{'})$. So let us assume the following condition:

Condition 1: $$ p_n^{5}-p_n = 5K^{'} \lt K_2 = K^8$$

So from now on, the partial refinement will be valid only those primes $p_n$ whose value $p_n^{5}-p_n$ is bounded by the eighth power of Ingham's $K$ constant.

4. Def. $K_3 = K_2^2$ an even bigger new fixed constant $K_3 \gt K_2$:

$$(p_{n+1} - p_n)^8 \lt K_3 + K_2 p_n$$

5. As $K_3 \gt K_2$, it is also true that:

$$(p_{n+1} - p_n)^8 \lt K_3 + K_2 p_n \lt K_3 + K_3 p_n = K_3 (1+p_n) \lt K_3 (2p_n) = (2K_3) \cdot p_n$$

6. Def. $K_4 = 2K_3$ as an even bigger new fixed constant $K_4 \gt K_3$:

$$(p_{n+1} - p_n)^8 \lt K_4 \cdot p_n$$

7. Now we make again the $8^{th}$ root in both sides:

$$p_{n+1} - p_n \lt K_4^{\frac{1}{8}} \cdot p_n^{\frac{1}{8}}$$

8. Finally, def. $K_5 = K_4^{\frac{1}{8}}$ as a new fixed constant. In this case $K_5 \lt K_4$ but still fixed and positive:

$$p_{n+1} - p_n \lt K_5 \cdot p_n^{\frac{1}{8}}$$

Only valid under Condition 1:

$$ p_n^{5}-p_n \lt K^8$$

being $K$ the original Ingham's $K$ constant.

Thus, if the manipulations and redefinition of constants are correct:

1. If $ p_n^{5}-p_n \lt K^8$:

$$p_{n+1} - p_n \lt K_5 p_n^{\frac{1}{8}}$$

for a fixed constant $K_5 = K_4^{\frac{1}{8}}=(2K_3)^{\frac{1}{8}}=(2K_2^2)^{\frac{1}{8}}=(2(K^8)^2)^{\frac{1}{8}}=2^{\frac{1}{8}}K^2$, and

2. In the rest of unbounded cases remains as the original Ingham's expression:

$$p_{n+1} - p_n \lt K p_n^{\frac{5}{8}}$$

I would like to ask the following questions:

1. Are the manipulations so far correct? It is based on Fermat's Little Theorem and the redefinition of the fixed constant.

2. If it is initially correct, is the refinement significant? It is bounded, so just valid for a finite range of prime numbers $p_n=2,3,5..$ as long as $ p_n^{5}-p_n \lt K^8$ but it could help to verify other open questions at least for the bounded range of primes covered by the refinement.