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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!

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  • $\begingroup$ you just finsh since $2^{k-1}\times (k+1) \geq 2^{k-1}\times2 = 2^k$ hence you prooved that $(k+1)! \geq 2^{k}$ $\endgroup$ – ALG Oct 23 '18 at 20:45
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Since $k+1\geqslant2$, $2^{k-1}(k+1)\geqslant2^{k-1}\times2=2^k$,

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What to do next:

$$k+1\ge2\implies 2^{k-1}(k+1)\ge2^{k+1-1}.$$

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