How to solve $(x!)!+x!+x=x^{x!} $

Question

How to solve this equation

$$ (x!)!+x!+x=x^{x!} $$

The answer is $3$ . But I have no idea of how to solve it. Thanks for your time.

Answer 1

I'm going to show $(x!)!>x^{x!}$ for $x>3$, which is clearly enough to show there are no solutions with $x>3$.

$(x!)!$ is a product of $x!$ factors, among which are $1,2,\dots,x$ and $x^2+1,\dots,x^2+x$ (since $x!\geq x^2+x$ for $x\geq 4$), as well as $x!-2x$ other factors, each of which is at least $x$. Therefore $$(x!)!\geq 1(x^2+1)\cdot 2(x^2+2)\cdot\dots\cdot x(x^2+x)\cdot x^{x!-2x}>(x^2)^x\cdot x^{x!-2x}=x^{x!}$$ as we wanted.