# 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!

• you just finsh since $2^{k-1}\times (k+1) \geq 2^{k-1}\times2 = 2^k$ hence you prooved that $(k+1)! \geq 2^{k}$ – ALG Oct 23 '18 at 20:45

Since $$k+1\geqslant2$$, $$2^{k-1}(k+1)\geqslant2^{k-1}\times2=2^k$$,
$$k+1\ge2\implies 2^{k-1}(k+1)\ge2^{k+1-1}.$$