How to prove that $n!>2^n$ holds for all $n>3$? I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$

The number $n!$ is the product of the $n-1$ numbers $2,3,\ldots,n$, each of which is greater than or equal to $2$ (and all of them but $2$ is actually greater than $2$) and, since $n>3$, at least one of them is greater than or equal to $4$. So, $n!>2^{n-2}\times4=2^n$
