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Prove using mathematical induction that for all $n! \ge 2^{n-1}$

Base case, p(1), 1! >= 1

$p(n+1), n!(n+1) \ge 2^{n-1}(n+1) $

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    $\begingroup$ n+1 is at least equal to 2.....so expression on the right of your last inequality is larger than 2^n $\endgroup$ – Alex M. May 27 at 0:23
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    $\begingroup$ The statement can also be proven without using mathematical induction: just observe that $n!$ is the product of $n-1$ integers $\ge 2$ (namely, the integers from $2$ to $n$), and so is $\ge 2^{n-1}$. $\endgroup$ – Geoffrey Trang May 27 at 0:40
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    $\begingroup$ Does this answer your question? Prove the inequality $n! \geq 2^n$ by induction $\endgroup$ – Culver Kwan May 27 at 1:27
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It looks like you've basically gotten it. You have

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

Now for all $n\geq1$, $n+1 \geq 2$. Therefore, from above, we've shown that

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

This completes the proof that $\forall n\geq 1,\quad n! \geq 2^{n-1}$.

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