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Decide for which $n ∈ \mathbb N$ the inequality

$2^n <2!$

is true and prove your claim by induction.

If $ n = 1$ hence $ 2^1 \not<1!$

If $ n = 2$ hence $ 2^2 \not<2!$

If $ n = 3$ hence $ 2^3 \not<3!$

If $ n = 4$ hence $ 2^4 <4!$

So for $n \geq 4$, $2^n <n!$ is true. Let's prove it.

  1. Base case ($n=4$):

$2^4< 4!$

  1. Inductive step ($n \to n+1$):

$2^n <n!$ | let's multiply by 2 both sides to get:

$2^n \cdot 2 < 2 \cdot n!$

$2^{n+1} < 2 \cdot n!$

As $n \geq 4$ hence $(n+1)>2$, then:

$2^{n+1} < (n+1) \cdot n!$

It it correct?

dvdgrgrtt
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  • It seems correct, you may add that $(n+1)\cdot n!=(n+1)!$. Also the line $2^n \cdot < 2 \cdot n!$ does not make much sense and seems unnecessary. – Gary Nov 29 '23 at 00:19
  • It is correct. I would make things crystal clear by writing "Suppose $2^n<n!$ for some $n\ge 4$" or "If $n\ge 4$ and $2^n<n!$" at the beginning of the inductive step. – Taladris Nov 29 '23 at 00:21
  • There's a typo in the line which follows "let's multiply by $2$ both sides to get:" you left off the new factor of $2$ on the left. Otherwise, looks good. – lulu Nov 29 '23 at 00:23
  • There are tidier ways of doing this: your initial step only needs to check $n=1$ as $n\ge 1 \implies (n+1)\ge 2$ and so $2^n <n! \implies$ $2^{n+1} =2 \cdot 2^n < 2 \cdot n! \le (n+1)\cdot n!=(n+1)!$ i.e. $2^n <n! \implies 2^{n+1} < (n+1)!$ – Henry Nov 29 '23 at 00:23

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