Looking at it on Desmos, it appears to be about 5.29, but what does this number mean, and how can it be shown in a mathematical proof?
In response to comments:
x is a real number greater than 0
x! is expandable by the gamma function so x can be a non-integer (which it will end up being)
How do I show in a mathematical proof what value of x where x > 0 causes e^x = x! to be true?
From the last line of your question, it seems that you might be more interested in the existence of a point that satisfies $e^x=\Gamma(x+1)$ and not its actual value.
You can know that a point where $e^x=\Gamma(x+1)$ exists as follows: Consider the function $f(x)=e^x-\Gamma(x+1)$. Since both $e^x$ and $\Gamma(x+1)$ are continuous for $x>0$, $f(x)$ is a continuous function. When $x=1$, $f(1)=e-1$, which is positive since $e>1$. When $x=10$, on the other hand, $f(10)=e^{10}-10!$ is negative, which we can see as follows: since $e<3$, $f(10)<3^{10}-10!$. We observe that $1<2$, $3<4$, $3<5$, $3<6$, $3<7$, $3<8$, and $3^2<10$. Therefore, \begin{align*} 10!&=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\\ &>1\cdot 1\cdot 3\cdot 3\cdot3\cdot3\cdot3\cdot3\cdot3^2\cdot3^2=3^{10}. \end{align*} Therefore, by the intermediate value theorem, there is some $x$ between $1$ and $10$ so that $f(x)=0$. Now, with better approximations (or Newton's method) you can approximate this point quite well.
I am not sure that the point $5.29\dots$ is really interesting on its own.