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I know that $\lim \sqrt[n]{n!} = \infty$, so, letting $a \in \mathbb{R}$, be fixed, I'm trying to get $n_0 \in \mathbb{N}$ such that $a^n < n!$ whenever $n>n_0$, but I can't figure it out.

I've been able to find a lower bound (possibly not too accurate), by AM-GM, but I can't conclude. Can someone give me a hint?

Gary
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manbolq
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    Hint: $$ e^n = 1 + n + \ldots + \frac{{n^n }}{{n!}} + \ldots > \frac{{n^n }}{{n!}} \Longrightarrow n! > \left( {\frac{n}{e}} \right)^n $$ – Gary Mar 23 '22 at 11:53
  • Certainly $100^{100}<100!$, so there's an upper bound on $n$. Factorials are tabulated at https://oeis.org/A000142/b000142.txt and it should be pretty easy to answer the question by staring at that list. – Gerry Myerson Mar 23 '22 at 11:55
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    @GerryMyerson Your inequality is the wrong way round. You have a strict lower bound on the minimum suitable $n$. – J.G. Mar 23 '22 at 11:58
  • @Gary Thanks, I can find such $n_0$ with that. However, how can I find the lowest $n_0$? – manbolq Mar 23 '22 at 12:01
  • @J.G., you're right, of course. – Gerry Myerson Mar 23 '22 at 12:01
  • $\left\lceil {e\cdot a} \right\rceil$ will be very close to the optimal $n_0$. – Gary Mar 23 '22 at 12:04
  • Use the Stirling approximation to find the smallest solution. – Peter Mar 23 '22 at 12:04
  • Somewhat related: https://math.stackexchange.com/questions/6581/how-to-prove-an-n-for-all-n-sufficiently-large-and-n-leq-nn-for-al and https://math.stackexchange.com/questions/994010/prove-by-induction-that-7n-n-for-all-integer-n-ge-21 – Gerry Myerson Mar 23 '22 at 12:05
  • A better approximation is $n_0 = \left\lceil {\frac{1}{2}W!\left( {\frac{{e^{2ae} }}{\pi }} \right)} \right\rceil$ which is probably optimal. Here $W$ is the Lambert-$W$ function. – Gary Mar 23 '22 at 12:08
  • Also https://www.quora.com/How-can-you-solve-n-100-n – Gerry Myerson Mar 23 '22 at 12:10
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    @GerryMyerson The question in the body is more general than the one in the title. Depending on what the OP actually want, it may not be a duplicate. – Gary Mar 23 '22 at 12:23
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    @Gary, I can't read the mind of OP to know which question OP really wants, but I would hope that OP could work out the more general answer from the specific one. – Gerry Myerson Mar 23 '22 at 12:27

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