How would you prove that $2^{n-1} > n!$?

Question

How do i prove that $2^{n-1} < n!$ for all $n \ge 1$

This is my proof:

Base case: Let n=1 then $2^{1-1} =1$ is the same on the right side so it holds

Inductive step: let $k \le 1$ we assume that the proposition holds for n=K

i know i want to show that $2^n < (n+1)!$

then i have $2^{n-1}= 2^n\times2^{-1} < n! \times 2^-1$ by inductive hypothesis but then i get stuck here.

Answer 1

$$n!=1\cdot 2\cdot 3\cdots n\ge 1\cdot 2\cdot 2\cdots 2=2^{n-1}$$ and the inequality is strict for $n\ge 3$. So actually the opposite of your inequality holds.

Answer 2

I think induction is an overly formal strategy here. The number $n!$ is a product of $n-1$ positive integers, each at least two. The number $2^{n-1}$ is also a product of $n-1$ positive integers, all of which are exactly two.