I want to show that $2^{k-1}\leq k!\hspace{0.2cm} $ $\forall \hspace{0.2cm} k \in \mathbb{N}$.
I tried this:
For $n=1$, $2^{0}=1=1!$, the equality holds. Suppose that holds for n, then
$2^{k}\leq(k+1)!$
$2^k\cdot2=2^{k+1}\leq(k+1)!\cdot2\leq(k+2)!$
So, the inequality holds for every $n \in \mathbb{N}$.
Is this right?