Proof by Induction for $n! > n2^n$

Question

For all natural number $n\in\mathbb{N}$ with $n\ge6$. Prove by induction that $n! > n2^n$. I proved the base step by showing that $n=6$ and that $720>384$. Then I assumed that $n=k$. Then for the third step I wanted to prove $n=k+1$ so I ended up with $(k+1)! \gt (k+1)2^{k+1}$. I'm not too sure how to continue on with the proof from here.

Answer 1

Note that $(k+1)! = (k+1)k!$. Now apply the induction hypothesis to the $k!$ term, together with the fact that $k>2$.

Answer 2

If $n \geq 6$ is such that $n! > n2^{n}$, then $(n+1)! > (n+1)n2^{n} > (n+1)2\cdot 2^{n} = (n+1)2^{n+1}$.