Factorial last AW

Looking at the graphs of $(x!)!$ and $x^{x!}$ (shown above), it appears that $(x!)!$ grows faster than $x^{x!}$ (I was too lazy to formally write a proof :) ). Therefore, as $x$ tends to infinity, $(x!)!$ should be much larger than $x^{x!}$ . Because of this, $(x!)! + x! + x$ should also be much larger than $x^{x!}$ for large values of $x$ .

Through trial and error, we can show that $0$ , $1$ , and $2$ do not, but $3$ does. However, once we try $4$ and $5$ , both graphs start growing rapidly (also shown in the graph above). Because $(x!)! + x! + x$ grows faster than $x^{x!}$ , they will never be equal again.

Therefore, $\boxed{3}$ is the only answer.

Let $f(x)=(x!)! +x! +x-x^{x! }$

Then $f(1)=2>0,f(2)=2>0,f(3)=0$

So $x=\boxed 3$ .