where does $x!$ overtake $e^x$? I basically want to find $x > 0$ such that $x! = e^x$ which turns into $ln(x!) = x$ 
But I'm not sure how to deal with the $ln(x!)$ on the left hand side.  I know the factorial function is defined for real numbers by the Gamma distribution, but I admittedly know very little about this. Is this even solvable?   
 A: If you generalize $x!$ to $\Gamma(x+1)$ (if $x$ is a natural number, then you have equality), then obviously for the smallest  $x_0$ there is the  symbolic solution $\Gamma(x_0+1) = e^{x_0}$. So the answer to the question is   $x > x_0 = 5.290316$ where the value for $x_0$ was obtained numerically by MATLAB.
A: We can actually prove the following statement by induction on n.
$$
e^n<n!
$$
for $n \geq 6$.
We start with the base case 6! is way bigger then $e^6$.
Then we assume that $e^k<k!$ for all $k \geq 6$.
We then wish to show $e^{k+1} <(k+1)!$
using properties of these functions we get
$$
ee^k<(k+1)k!
$$
then
$$
e^k<\frac{k+1}{e}k!
$$
And we know that $e^k$ is always less then $k!$ by inductive assumption  (also note  $\frac{k+1}{e} >1)$ so we are done.
A similar argument could be made in a similar way for the gamma function?  Perhaps??? 
A: It's easy enough to check that
$$5!<e^5$$
$$6!>e^6$$
Thus, the solution must lie in $5<x<6$.  We generalize the factorial with the Gamma function to end up with the following line:
$$\Gamma(x+1)=e^x$$
Take the log of both sides to get
$$\ln(\Gamma(x+1))=x$$
$$0=\ln(\Gamma(x+1))-x$$
Applying Newton's method to get
$$x_{n+1}=x_n-\frac{\ln(\Gamma(x_n+1))-x_n}{\psi(x_n+1)-1}$$
where $\psi(x)$ is the digamma function.  With $x_0=5.5$, I get
$$\begin{array}{c|c}n&x_n\\\hline0&5.50000\\1&5.29498\\2&5.29032\\3&5.29032\end{array}$$
which is the solution out $5$ places.
