Proving $ n! \geq 2^{n-1} $

Question

Prove that $$ n! \geq 2^{n-1}$$ for $n \geq 1$.

My initial solution by induction goes like this.

For $n = 1 : 1 \geq 1 $.

Assuming that $$ n ! \geq 2^{n-1}.$$ Then for $n+1$, $$ (n+1)! = 2^{n+1-1} $$ so

$$ n!(n+1) = 2^{n-1} \cdot 2 $$ How I can finish?

Answer 1

Use your induction hypothesis and $n+1>2$.

Answer 2

Hint: $(n+1)! = (n+1) \cdot n! \geq (n+1) \cdot 2^{n-1} \geq 2 \cdot 2^{n-1} = 2^{(n +1) -1}.$

Answer 3

Why use induction here? Just consider $\frac{1 \cdot 2 \ldots n}{2 \cdot 2 \ldots 2}$ There are $n-1$ terms in the numerator after the first one and $n-1$ in the denominator. Clearly the numerator is monotone increasing and $k>2 \ \forall \ k \geq 2$, so the fraction is greater than 1.