Decide for which $n ∈ \mathbb N$ the inequality
$2^n <2!$
is true and prove your claim by induction.
If $ n = 1$ hence $ 2^1 \not<1!$
If $ n = 2$ hence $ 2^2 \not<2!$
If $ n = 3$ hence $ 2^3 \not<3!$
If $ n = 4$ hence $ 2^4 <4!$
So for $n \geq 4$, $2^n <n!$ is true. Let's prove it.
- Base case ($n=4$):
$2^4< 4!$
- Inductive step ($n \to n+1$):
$2^n <n!$ | let's multiply by 2 both sides to get:
$2^n \cdot 2 < 2 \cdot n!$
$2^{n+1} < 2 \cdot n!$
As $n \geq 4$ hence $(n+1)>2$, then:
$2^{n+1} < (n+1) \cdot n!$
It it correct?