Equations involving factorial/Gamma function 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.
 A: 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.
A: 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.
