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How would I go about proving that $n! \ge 2^{n-1}\ \forall n \ge 1$? The base case makes sense to me, but when I do the inductive step, I go here using the inductive step:

$$ n+ 1 = k+1 $$ $$ (k+1)! \ge 2^{(k+1)-1} $$ $$ (k+1)! = (k+1)\cdot k!$$ $$\ 2^{k-1+1} = 2^k = 2^{k-1} \cdot 2 $$ $$(k+1) \cdot k! \ge 2^{k-1} \cdot 2 $$ $$ \therefore (n+1) \cdot n! \ge 2^{n-1} \cdot 2$$

Is this a viable proof? If not, what do I need to do differently in order to make it correct? Have I not gone far enough?

apizzimenti
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  • See http://math.stackexchange.com/q/1439081 –  Nov 11 '15 at 00:19
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    You just need $(k+1)!=(k+1)k!\geq (k+1)2^{k-1}\geq 2\cdot 2^{k-1}=2^k$ – Alex Fish Nov 11 '15 at 00:20
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    Personally I would say that if it were true for $n=k$ then $k! \ge 2^{k-1}$ and so $(k+1)! = (k+1)k! \ge 2k! \ge 2\times 2^{k-1} = 2^{(k+1)-1}$, i.e. it would be true for $n=k+1$. The way you have written is has what I think you are trying to show as the second line – Henry Nov 11 '15 at 00:20

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Assuming your result holds for $n=k$, i.e. $$k! \geqslant 2^{k-1},$$ you then have $$(k+1)! = (k+1) \times k! \geqslant (k+1) \times 2^{k-1} \geqslant 2^k,$$ where the first equality is the definition of the factorial, the second is the case $n=k$, and the third is true because $k\geqslant1$.

onamoonlessnight
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