# Proving $n! \ge 2^{n-1 }$for all $n\ge1$by mathematical Induction [duplicate]

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Im trying to solve the following question In the second step where do they get $k!=2^k-1?$

## marked as duplicate by GNUSupporter 8964民主女神 地下教會, Hans Engler, B. Mehta, HK Lee, Chris CusterMay 11 '18 at 10:59

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• They never wrote $k! = 2^k - 1$, or even $k!=2^{k-1}$. They only used the induction hypothesis $k!\geq 2^{k-1}$. – 5xum Sep 17 '15 at 5:51
• They wrote $k! = 2^{k-1}$, which is the induction hypothesis. – Christopher Carl Heckman Sep 17 '15 at 5:52
• @Yash Malik can you fix your edit please – techno Sep 17 '15 at 6:06
• @Yash Malik With due respect,Please stop editing if you dont know what you are doing. – techno Sep 17 '15 at 6:15
• @YashMalik Thanks :) its okay now.No probs. – techno Sep 17 '15 at 6:17

## 2 Answers

In the induction step, they say "Suppose that $k! \ge 2^{k-1}$ for some $k \ge 1$."

That then allows them to say "$(k+1)\,k! \ge (k+1)\,2^{k-1}$", multiplying both sides of the inequality by the positive $(k+1)$.

• @Henry Do you think you could correct the mess there? – Did Sep 17 '15 at 8:40

Consider this example...similar to your problem..

Prove $n!>2^n$ $(n \geq4)$ Prove it at first for any $n \geq 4$.

Assume it true for some $k$. let it be true $k!>2^k$

Now we try to prove it true for $k+1$. We have to prove $(k+1)!>2^{k+1}$.

We know $2^{k+1}$=$2^k*2$.

Also,we know $(k+1)!=k!*(k+1).$ We also assumed earlier $k!>2^k$.

So, we can write $k!(k+1)>2^k*2$ [Since,$k>4 ,k+1>4>2$ which is in the RHS and $k!>2^k$ as we assumed earlier].

Hope this helps you to understand the problem and solve it in a similar way.

• @techno-Welcome!!Hope this helps you!!. – tatan Sep 17 '15 at 6:52
• @techno-Please upvote if it helps :-)! – tatan Sep 17 '15 at 7:00