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Mathematical Induction:-Prove that $n! \geq 2^{(n-1)}$ for $n\geq 1$.

I tried mathematical induction but could not

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marked as duplicate by Andrey Rekalo, Start wearing purple, Davide Giraudo, MathOverview, Git Gud Aug 8 '13 at 11:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you search this site a little, you will find several very similar questions, for example, Proof the inequality $n! \geq 2^n$ by induction and Is this induction procedure correct? ($2^n<n!$). – Martin Sleziak Aug 8 '13 at 8:22

Base case: If n = 1, then $1 = 1! \geq 1 = 2^{1 - 1}$.

Induction step: Suppose that $n! \geq 2^{n - 1}$ for some $n \geq 1$; we must show that this holds if we replace $n$ by $n + 1$. Now we have $n + 1 \geq 2$, so

$$(n + 1)! = (n + 1) n! \geq 2 (n!) \geq 2(2^{n - 1}) = 2^{n} = 2^{(n + 1) - 1}$$

as desired. Note that the second inequality is where we use the inductive hypothesis.

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It's not that hard. First we show that if $n=1$ the claim holds: $$n! =1! =1 = 2^0 =2^{1-1}$$ Now suppose that there exist $n \in \mathbb{N}$, so that your statement holds. Then $$(n+1)!=(n+1)\cdot n! \geq (n+1) \cdot 2^{n-1}$$ The inequality above follows by your induction hypothesis. Now $$(n+1)! \geq n\cdot 2^{n-1} + 2^{n-1} \geq 1\cdot 2^{n-1} + 2^{n-1} =2\cdot 2^{n-1}=2^n$$ And this completes your proof. Is this clear?

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If $n=1$, clearly equality holds.

If $n \geq 2$, $n!=n(n-1)(n-2) \dots(2)(1)$ has the first $n-1$ factors $\geq2$, and the last factor $1$.

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