Prove by induction that $$ (n!)^2 \geqslant n^n$$ for all positive natural numbers.
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If you do not now how to go about trying an induction proof, there are many helpful resources online. Even the wikipedia page about induction gives the standard way of going about it. It would help if you attempted this question, then you can edit your post with your attempt if you get stuck, and we will help you! – astromoner Dec 07 '19 at 21:47
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Hint: Use this – Maximilian Janisch Dec 07 '19 at 21:55
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Near-duplicate of Prove that $n! \geq n^{\frac{n}{2}}$ – Maximilian Janisch Dec 07 '19 at 21:56
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Note that $$(n+1)^{n+1}/n^n=(n+1)\cdot\left(\frac{n+1}n\right)^n$$$$=(n+1)\cdot\left(1+\frac1n\right)^n<(n+1)\cdot e$$
And, $$(n+1!)^2/(n!)^2=(n+1)^2$$
So, $(n+1)^2>(n+1)\cdot e$ when $n+1>e$, which happens when $n>1$. Can you take it from here?
Rushabh Mehta
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