@ErickWong , you are correct about the approximation error due to unconverted base, but that can be corrected easily by base change formula, are you saying that: after converting the base , the result I claim will be wrong ?

Also, would you please, elaborate your comment " The argument in math.stackexchange.com/questions/1892924 is essentially circular so it can't prove anything substantial. " . Thanks for your help.

Once you correct the error in approximation, no contradiction will be possible without introducing some new idea. Your current argument boils down to applying Stirling's approximation (in $\log(n!)$ form) and then comparing it to Stirling's approximation (in $\sqrt{2\pi n} (n/e)^n$ form). This is circular: when one of the two Stirling's was transcribed incorrectly, there was a specious contradiction;

ah... now I understand what you mean by "circular".

But if all you are doing is expanding out Stirling's approximation, then nothing good can come out of it because you have essentially removed all but a sliver of the number-theoretic content. There are obviously real-valued solutions to $k^2 = n! + 1$, and you haven't yet done anything substantial to capture the unusualness of $k$ and $n$ being integers.

@ErickWong my idea was/is to make the equation a one variable equation, then see how big or small it is compare to n!... what about this approach?

If size alone was enough to do the trick, then there wouldn't be real-valued solutions.

then my idea seems to be fundamentally inaccurate!

I would guess that the perfect squares are spaced closely enough together to well-approximate $n!$ up to the error term of Stirling's infinitely often.

What you think of equation 1.2 of my answer? Don't you think, this is unusaul to have such property?

@ErickWong , It seems quite counter intuitive to me, that even if the right hand side of equation 1.8 (referring to my answer) is greater than $n^n$ , it is not enough, don't you think so? I would learn a lot if you give example of such phenomenon , i.e. example of "size alone was enough to do the trick, then there wouldn't be real-valued solutions."

Is everything ok in equation 1.2 (referring to my answer to the question $n!+1$ being a perfect square)?

@ErickWong , I was expecting your comment on eqaution 1.2 of my answer :) (referring to my answer to the question $n!+1$ being a perfect square)

In 1.2, you define $k=4t-1$, then you rewrite $4t-1 = a$. Can you see that all you are doing is manipulating in circles with no progress? There is nothing useful about 1.2. After you fix equation 1.8, the $(e/2)^n$ term will become $(e/e)^n$ or $(2/2)^n$ and you will not have a contradiction. Which is as it should be because Stirling's formula is internally consistent.

@ErickWong Would you suggenst me a note/book to get good idea in number theoretic-approximation, please?

@ErickWong I might be wrong but I think, After $(e/2)^n$ term will become $(e/e)^n$ or $(2/2)^n$ , then LHS will have $2^{O(\log(n))}$ and RHS will have $\sqrt n$ (seeing the equation 1.8), I will do the calculation in detail .

I really appreciate your help. Thanks! If possible, please suggenst me a note/book to get good idea in number theoretic-approximation.

$2^O(\log n)$ is literally the same as $n^{O(1)}$, which matches a great number of functions: the RHS could be $\sqrt n$, or $100000/n^{100}$ or even $\exp(-\sqrt{n}) + n^{1 + \sin n}$, there would be no contradiction.

It wouldn't hurt for you to read (and do exercises) from Tom Apostol's Introduction to Analytic Number Theory, say Chapter 3. However, be aware that it is highly unlikely anything so elementary will lead to a proof of Brocard's conjecture (in most cases if that were possible then Erdős would have already worked it out :).