Are there any known methods to formally solve equations like:
1)$x^3!+(2x^2)!-x!+3=0$
2)$x!=e^x$ ($0$ is trivial but there must be another one)
3)$(2x!)^2+x!-1=0$
4)$x!!+x!=7$
I don't need approximation (which are not so hard to find) but I mean closed forms, I think that there could be a lot of more equations of this kind but these are the first that came into my mind.
On
To some extent it seems like you'll need relationships between, for example, $x^2!$ and $(x!)^2$.
That might be a tough nut to crack, or not. I tried a few and any pattern I could see broke down quickly:
$$0^2! = (0!)^2 \\ 1^2! = (1!)^2 \\ 2^2! = (2!)^2 \cdot 3! \\ 3^2! = (3!)^2 \cdot \frac{8!}{4!} \\ 4^2! = (4!)^2 \cdot \frac{15!}{6!} \cdot 5 \cdot 4 \\ 5^2! = (5!)^2 \cdot \frac{24!}{8!} \cdot 7 \cdot 5 \cdot 2$$
There may be a closed form hidden in there somewhere, but I can't see it (or cannot find reference to it).
This is just one kind of relationship that should be determined in order to attack some of the problems you've proposed.
I'm able to solve (2) and (3). (2) for its trivialness and (3) for its generalization properties. $$x!=e^x$$ Since, $$0!=1=e^0$$ $0$ is a solution to this equation. $$(2x!)^2+x!-1=0$$ Set $u=x!$ and then solve the resulting quadratic equation... $$4u^2+u-1=0$$ $$u=x!={{\pm \sqrt{17}-1} \over 8}$$ You mentioned that the ! could be interpreted as the gamma function, so lets do so... $$\Gamma (x+1)={{\pm \sqrt{17}-1} \over 8}$$ Since almost all evaluations needing the gamma function are numerical, this is a proof that you have to use numeric methods. Your solution is... $$x=\Gamma^{-1} \left({{\pm \sqrt{17}-1} \over 8}\right)-1$$ where $\Gamma^{-1} (t)$ is the inverse gamma function.