Question : What is the inverse factorial of $3$? Basically, if $n! = 3$, then what is $n$ equal to?
Well, if you look at every factorial, not one of them equals $3$. So how about using the gamma function instead? Well, it probably works. I haven't tested it out yet but it would probably be an irrational number.
But every single time I search up inverse factorial it always says that the reverse factorial or inverse gamma function cannot exist because there is no gamma function or factorial where $n! = 3$ etc.
So, would it be possible for the inverse factorial of $3$ to exist, but just as an irrational number?
A few things. First off... formally factorials may only take natural number inputs. Trying to talk about $(2.4059)!$ is not allowed. If we want to talk about something like that we are no longer talking about the factorial function, but rather the gamma function. In that regard, no... there does not exist an "inverse factorial." There does not exist any natural number value of $n$ such that $n!=3$.
Now... the gamma function is defined in such a way as to have many of the same properties as the factorial function and even agree with the factorial function for all of its natural number inputs (after a slight offset: $\Gamma(n+1)=n!$) as well as to have several other desirable properties (e.g. it is "analytic")... but it needs to be emphasized that this is just one of many such functions who have these properties. It happens to be the most common and most popular one, but it is by no means the only one.
The gamma function does have some real values $x$ such that $\Gamma(x)=3$, but it has several. To be an "invertible function" we want two things... For each $y$ in our codomain, we want at least one value of $x$ in our domain such that $f(x)=y$... a property called surjectivity. We also want for each $y$ in our codomain, we want at most one value of $x$ in our domain such that $f(x)=y$... a property called injectivity. Here, the gamma function fails at both. There does not exist any value of $x$ such that $\Gamma(x)=0$, and for things like $\Gamma(x)=3$ there exist several values of $x$ doing this, including but not limited to values near $3.4, 0.3, -1.3, -1.8, -3,\dots$. For this reason, we say that there does not exist a true inverse gamma function.
This is similar to how you will have heard that if we want to solve $x^2=9$ for $x$, we actually get more than one solution. We get both $3^2=9$ and $(-3)^2=9$. That being said, just like we did for square roots where we can define the symbol $\sqrt{~~}$ to mean that we want the positive (and only the positive) answer to the question of what value of $x$ satisfies $x^2=9$, we can do similar here. That is what we mean by "choosing a branch." We can specify that when asking what value of $x$ satisfies $\Gamma(x)=y$ we want only the value bigger than $1$ if it exists. That way, we can get what we would have otherwise expected... that $\Gamma^{-1}(6)=4, \Gamma^{-1}(24)=5, \Gamma^{-1}(120)=6$ and so on. If we were to use this meaning and definition of $\Gamma^{-1}$, then yes we do have an answer to the value you are interested in. Yes, it will be irrational. It will be a value close to $2.4059\dots$ as you should have guessed since $2!=2<6=3!$ so we would have expected the value to be between $2$ and $3$.
No, there doesn't exist a very clean way to write the value in any way you are already used to... we would just write the value as $\Gamma^{-1}(3)-1$... effectively saying the answer is "the answer." We can use numerical methods to try to figure out the digits of the number to see where it is close to, but there will not exist an exact value that we can find for most cases.