Prove that $n! \geq 2^{n-1}$ for $ n\geq1$ [duplicate]

Solution 1:

Base case: If n = 1, then $1 = 1! \geq 1 = 2^{1 - 1}$.

Induction step: Suppose that $n! \geq 2^{n - 1}$ for some $n \geq 1$; we must show that this holds if we replace $n$ by $n + 1$. Now we have $n + 1 \geq 2$, so

$$(n + 1)! = (n + 1) n! \geq 2 (n!) \geq 2(2^{n - 1}) = 2^{n} = 2^{(n + 1) - 1}$$

as desired. Note that the second inequality is where we use the inductive hypothesis.

Solution 2:

It's not that hard. First we show that if $n=1$ the claim holds: $$n! =1! =1 = 2^0 =2^{1-1}$$ Now suppose that there exist $n \in \mathbb{N}$, so that your statement holds. Then $$(n+1)!=(n+1)\cdot n! \geq (n+1) \cdot 2^{n-1}$$ The inequality above follows by your induction hypothesis. Now $$(n+1)! \geq n\cdot 2^{n-1} + 2^{n-1} \geq 1\cdot 2^{n-1} + 2^{n-1} =2\cdot 2^{n-1}=2^n$$ And this completes your proof. Is this clear?