Prove by the method of finite induction

Question

Using the method of finite induction show that $2^{(n-1)} \leq n! \forall n$ belong to $N$ (Natural Number) Can someone please prove this?

Answer 1

For $n=0$ and $1$, it is true.

let $n\geq 1$ such that

$2^{n-1}\leq n!$.

$\implies$

$2^n=2.2^{n-1}\leq 2.n!\leq (n+1).n!$

$\implies$

$2^n\leq (n+1)!$.

thus

$\forall n \in \mathbb N \;\; 2^n\leq n!$.