Prove by induction, for all positive integers $n$, that $n!\ge2^{n-1}$ So far I have:
Base case: $n=1$: LHS: $1!=1$, RHS: $2^{1-1}=1$, $1\ge1$ so $P(1)$ is true.
Inductive step: Assume $n=k$ is true, that is $k!\ge2^{k-1}$, we must show that $n=k+1$ is true, that is $(k+1)!\ge2^k$.
$(k+1)!=k!(k+1)\ge2^{k-1}(k+1)$ (since we assume $k!\ge2^{k-1}$)  

Not really sure what to do after this, I know that $2^{k-1}=$$2^k\over2$ but I don't know if that helps at all.  

Thanks in advance for any help!
Since $k+1\geqslant2$, $2^{k-1}(k+1)\geqslant2^{k-1}\times2=2^k$,
Since $k+1\geqslant2$, $2^{k-1}(k+1)\geqslant2^{k-1}\times2=2^k$,