Let $x \geq 0$. I'm trying to prove that $x^n \leq n!$ for all natural numbers $n \geq xe$. Stirling's approximation suggests that this inequality holds; indeed $$ n! \approx \sqrt{2\pi n} (n / e)^n \geq x^n \quad \text{for all $n \geq xe$,} $$ but I'm not sure how to make this rigorous. Also, it would be best if this could be proven without needing to use Stirling's approximation.
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2 Answers
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The only case of the problem you really need to solve is the extreme case where $n=xe$. Hence, you want to show that: $$\bigg(\frac{n}{e}\bigg)^n \leqslant n!$$ Can you proceed from here? You could try using induction.
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It's possible to prove the inequality mentioned in @Haran's answer using induction, but it also follows by writing the Taylor series expansion for $e^n$: $$ e^n = \sum_{k = 0}^\infty \frac{n^k}{k!} \geq \frac{n^n}{n!}. $$