TITLE: Proof of $n!\geq2^{n-1}$ by mathematical induction
QUESTION [6 upvotes]: I am trying to make a proof
$$n!\geq2^{n-1}\;\;\forall \;n\in N$$
Here's what I've done!
$ \text{When}\;\; n=1,\;\; 2^{0}\leq 1!$,
$ \text{when}\;\; n=2,\;\; 2^{1}\leq 2!$,
$ \text{when}\;\; n=3,\;\; 2^{2}\leq 3!$,
$\vdots$
Assume it is true for $n=k$, then
$$2^{k-1}\leq k!$$.
Now, we want to prove for $n=k+1$.
I got stuck at this point. I need help! Thanks!

REPLY [3 votes]: If $n=1$, we have $1!=1\ge 1= 2^0$.
Suppose that $n!\ge 2^{n-1}$ for some $n\ge1$.
Then $$(n+1)! = (n+1)\cdot n! \ge 2\cdot n! \ge 2\cdot 2^{n-1} = 2^n$$
So, by induction, $n!\ge 2^{n-1}$, for all $n\in\mathbb{N}$.