Do exponents have higher time complexity than factorials?
In fact $e^n$ is $o(n!)$. We could show it using the Stirling approximation by taking limit
$$\lim_{n\rightarrow \infty}{\frac{e^n}{(\sqrt{2\pi n})(\frac{n}{e})^n}} = \lim_{n\rightarrow \infty}{\frac{e^{2n}}{(\sqrt{2\pi n})n^n}} = \lim_{n\rightarrow \infty}{\frac{1}{\sqrt{2\pi n}}\frac{e^{2n}}{n^n}}$$
Since $\lim_{n\rightarrow \infty}{\frac{1}{\sqrt{2\pi n}}} = 0$ we only need to calculate
$$\lim_{n\rightarrow \infty}{\left(\frac{e^2}{n}\right)^n}$$
The function $\frac{e^2}{n}$ decreases faster than any function $c^n$ where $0 &lt; c &lt; 1$ for sufficiently large $n$s (as $n$ goes to infinity), and so $\lim_{n\rightarrow \infty}{\left(\frac{e^2}{n}\right)^n} = 0$.
Regarding Quazi Irfan's comments about Matlab graph
I am not a Matlab user, but these are plots for $e^x$ and Stirling approximations of $n!$ using Wolfram script: plot (1\sqrt(2*pi*x)*(x/e)^x) from 0 to 6, plot(e^x) from 0 to 6

As you see $n!$ overtakes $e^n$ when $n=6$.