About the unicity of the analytic continuations of $\zeta$ and the continuation used to have $\zeta'(0) = -(\ln(1) +\ln(2) +\ln(3) + \ldots )$ The analytic continuation of an analytic function is unique. I've come across different representations of those, for example: $\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s), s\neq1,$ AKA functional eq.
$\zeta(s)=\frac1{s-1}+\sum_{n=1}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n, s\neq1,$ from here, etc.
Now, on one hand if one takes the derivative of the functional equation and evaluates at $0,$ one gets $-\ln\sqrt{2\pi},$ e.g. proof here.
On the other hand, in this proof, is mentioned:

Now if you write formally the derivative of the Dirichlet-series for zeta then you have
$$ \zeta'(s) = {\ln(1) \over 1^s}+{\ln(1/2) \over 2^s}  +{\ln(1/3) \over 3^s} + \ldots $$
This is for some s convergent and from there can be analytically continued to $s=0$ as well from where the the formal expression reduces to
$$ \zeta'(0) = -(\ln(1) +\ln(2) +\ln(3)  + \ldots )$$
which is then formally identical to $ - \lim_{n \to \infty} \ln(n!)$ .

That is, one ultimately gets $ - \lim_{n \to \infty} \ln(n!)$
In that same proof, mentions that $\zeta'(0)=-\ln\sqrt{2\pi}$
Ultimately, one would get to "$\infty!=\sqrt{2\pi}$"
Why is that?
So far, what I understand is that if $f$ is unique, then regardless of how it's presented, say $f'$, then $f(x)=a$ should be equal to $f'(x)=a.$ But in this example looks like $f(x)=a$ and $f'(x)=b.$
I don't know much about this. Excuse my ignorance in advance :)
 A: $\infty!$ is not the same as $\sqrt{2\pi}$, and "proofs" that they are rely on a sleight of hand between functions as mathematical objects and informal notions of functions as something whose values you compute.
When we write something like "$f(z)=1/(1-z)$" we're suggesting to the reader a preferred way of computing the function's values, but that way of computing isn't part of the function. Stealing and adapting an example from the first question you linked, here are two representations of the same function from $\{z:|z| < 1\}$ to  $\mathbb{C}$. $$\frac{1}{1-z}=f(z)=\sum_{n=0}^\infty \frac{1}{z^n} \text.$$
In the next few sentences I'm going to refer to $f_a$ to mean "the function $f$ when given the representation $1/(1-z)$" and $f_b$ for "$f(z)$ when given the representaiton $\sum_0^{\infty}1/z^n$," but I don't want to let something I'm doing for notational convenience confuse the point: there is exactly one function under discussion. If you put $i/2$, into $f_a$ you'll get $(4+2i)/5$. If you put $i/2$ into $f_b$ you'll get $(4+2i)/5$, and so it goes with any number in the domain. Having the same domain, and putting the same number in and getting the same result out make $f_a$ and $f_b$ the same object.
But this doesn't make the methods of calculating associated to $f_a$ and $f_b$ the same. The fact that these methods of calculating are not the same is one of the primary motivations for looking for analytic continuations in the first place. Suppose we only knew about $f$ through the calculation suggested by $f_b$. We've restricted the domain of $f$ to $\{z : |z|<1\}$ by fiat for this example, but the infinite series approach to finding values of $f$ presents a bigger obstacle: the series diverges, so our way of calculating $f_b$ does not produce meaningful answers outside of its original domain. We can find another function $g:\mathbb{C}\setminus1\rightarrow\mathbb{C}$ with a broader domain that acts the same as $f$ on $f$'s domain, specifically $g(z)=1/(1-z)$,* but it makes no sense to say that $g$ is calculated by summing the series $\sum_{n=0}^\infty z^{-n}$. If this were a viable way of calculating $g$ we wouldn't have had any reason to find a new representation when extending $f_b$ in the first place!
This confusion between functions as objects that associate numbers in their domain to other numbers in their range is what makes the pseudo-proof of $\infty!=\sqrt{2\pi}$ "work." Taking a look at the answer to the question you linked, we see this:

Now if you write formally the derivative of the Dirichlet-series for zeta then you have $$\zeta'(s)=\frac{\ln(1)}{1^s}+\frac{\ln(2)}{2^s}+\frac{\ln(3)}{3^s}+\dots$$ This is for some $s$ convergent and from there can be analytically continued to $s=0$.

That we can find an analytic continuation of $\zeta'$ outside of the domain $\text{Re}(z)>1$ to everywhere in the complex plane but $s=1$ means we can find exactly one function $Z'(s)$ where $Z'$ is differentiable and $Z'(s)=\zeta'(s)$ everywhere that $\zeta'$ is defined. It does not mean that the values of $Z'$ can be reasonably computed by the series $\sum_{n=1}^\infty n^{-s}\ln(n)$. What we are continuing is the function, not some preferred means of computing it, and again, if we could "continue" the series we really wouldn't need to think much about finding analytic continuations at all, it would just be obvious.
It is true that $Z'(0)=-\ln\sqrt{2\pi}$, and it is true that the given way of computing $\zeta'$ as $\sum_{n=1}^\infty n^{-s}\ln(n)$ produces a series that diverges to $\infty$ at $0$ by way of $\lim\ln(n!)$, but this does not make $\infty!$ equal to $\sqrt{2\pi}$.
*Which is not the same as $f_a$ due to different domains of defintion, so really there are two sleights of hand at work here.
