Does $y=\lim_{n\to\infty}x\uparrow\uparrow n-f(x,n)$ have a root at $2$?

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While messing around on Desmos calculator, I found an interesting factorial/tetration function, where we are using Knuth arrow notation.

$$y=\lim_{n\to\infty}x\uparrow\uparrow n-f(x,n)\\f(x,n)=x\underbrace{!\dots!}_n$$

Where there are $n$ factorials following $x$ in $f(x,n)$.

Looking at it graphically, you may find the link above, it appears there are two roots, $x=1,a$, where $a$ seems to get increasingly closer to $2$ as $n$ goes to infinity.

However, I can't figure out how to prove this because limits like this are not the sort I can deal with.

I do notice, however, that the following is the infinite power tower:

$$g(x)=x^{x^{x^{x^{\dots}}}}$$

And the inverse of the infinite power tower is:

$$g^{-1}(x)=\sqrt[x]x$$

So the real problem is probably in the my factorial function $f(x,n)$, which I find hard to deal with.

Any solutions or directions to point?

EDIT: It appears as though this root, $x=a$, will most likely occur higher than $2$, which is going to give us a problem. The infinite power tower doesn't hold for $x>\sqrt[e]e$, which means we might have to do this some harder way...