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$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$

Since $n\ge 4$ we have $n! = 1\cdot 2\cdot 3\cdot 4 \cdot 5\cdots n = 24 \cdot 5 \cdots n > 16 \cdot 5 \cdots n\ge  2^4\cdot 2 \cdots 2 = 2^42^{n-4}=2^n$
