User blog:DrCeasium/Hyperfactorial Notation - Take 2

After the not-quite-up-to-what-I-wanted attempt 1 of making a hyperfactorial notation, I have decided to go back to the drawing board and start again. This is what I've come up with:

Still sticking with factorial as it is one of the less developed area of googology, I have come up with a simple notation that does not involve arrays and is also not that similar to any of Hyper-E notation. It is relatively powerful and uses "!" as symbols. It also adds two features to the exclamation mark, a subscript and superscript, so the final symbol is in the form $$!_a^b$$. The b is just short for how many there are, so, for example $$!_a^4$$ could be written as $$!_a!_a!_a!_a$$. There are 4 rules to evaluate this notation:

For $$n!_a^b$$ where n, a and b are positive whole numbers (excluding 0), Here are two examples of how it works: As usual more to come, all comments are appreciated, named numbers soon if this notation actually works as I want it to, and if it does I might get my own website and put them on there.
 * 1) If there are multiple exclamation marks, always evaluate the one on the left (or the one closest to n)  first.
 * 2) If b > 1 and a > 1, $$n!_a^b = (n!_{a-1}^n)!_a^{b-1}$$
 * 3) If b > 1 and a = 1, $$n!_1^b = ((\cdots((n!_1^{b-1})!_1^{b-1})\cdots)!_1^{b-1})!_1^{b-1}$$ with n $$!_1^{b-1}$$'s. Note: you cannot just try and simplify this by adding together the powers of the exclamation marks: you would be at a higher power than at the start (usually). This is because (and why) there are brackets.
 * 4) If b = 1, $$n!_a^1 = n!$$. Note: this can be done to any exclamation mark, not just the one nearest n.
 * 1) $$4!_1^2$$- just a small one to get started. It = (by rule 2) $$(((4!_1^1)!_1^1)!_1^1)!_1^1$$ = (by rule 3) $$(((4!)!)!)!$$ = $$((24!)!)!$$ ~ $$((6.4\times10^{24})!)!$$. A helpful formula when dealing with factorials is $$n!\approx\sqrt{2\pi n}\times\left(\cfrac{n}{e}\right)^n$$ (see the wikipedia article here ). This can be used to give an approximation of this as $$(10^{2.5\times10^{24}})!$$ and a (very rough) approximation of this at $$10^{10^{10^{24}}}$$. Definitely "just a small one". Not like it's bigger than a googolplex or anything.
 * 2) $$3!_2^2$$ = (by rule 1) $$(3!_1^3)!_2^1$$ = (by rules 2 & 3) $$(((3!_1^2)!_1^2)!_1^2)!$$ = (by rule 2) $$(((((3!_1^1)!_1^1)!_1^1)!_1^2)!_1^2)!$$ = (by rule 3) $$(((((3!)!)!)!_1^2)!_1^2)! = (((720!)!_1^2)!_1^2)! \approx (((2.6\times10^{1747})!_1^2)!_1^2)!$$ = (by rule 2) $$((((\cdots((2.6\times10^{1747})!)!\cdots)!)!)!_1^2)!$$ with $$2.6\times10^{1747}$$ !'s. Then get that number (without the last two !'s) and call it x. Then you have x followed by x+1 !'s (+1 from the extra one on the end). Not bad.