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I've encountered a proof which claims $n! > n^2$ for sufficiently large $n$. I tried using induction to prove it for an arbitrary $a$: $n! > n^a$.

Lets assume the claim is true for $n$: $n! > n^a$. We need to prove it for $n+1$:

$$(n+1)! = (n+1)n! > (n+1)n^a \ge n^{a+1} + n^a$$

How to continue?

I thought about using the binomial formula for $(n+1)^a$:

$$ (n+1)^a = \sum\limits_{n=1}^{a}{ {n \choose k}n^k} = n^a\sum\limits_{n=1}^{a-1}{ {n \choose k}n^k} < n! \sum\limits_{n=1}^{a-1}{ {n \choose k}n^k}\dots$$

Elimination
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    See http://en.wikipedia.org/wiki/Stirling%27s_approximation. – user_of_math Aug 24 '14 at 10:46
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    Even when you succeed in showing that $n!>n^a$ implies $(n+1)! > (n+1)^a$, you still have to show that there exists some $n$ for which $n! > n^a$. – user133281 Aug 24 '14 at 11:54
  • And once you have shown that, your induction step is easy to complete: assume wlog $a>0.$ Then, for $n$ sufficiently large, $(n+1)n^a\ge(n+1)^a$ since $n+1\ge2^a\ge(1+1/n)^a.$ – Anne Bauval Feb 27 '23 at 11:02

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If $n > 2a$, it is not hard to prove that $n! > n^a$. Since $2(n-a)>n$ and $n-a+1 > a$, we have

$$n^a = n \cdot n \cdots n < n \cdot (2 (n-1)) \cdot (3(n-2)) \cdots (a(n-a+1)) \\ = 2 \cdots a \cdot (n-a+1)(n-a+2) \cdots n < n!$$

Dongryul Kim
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    Note that $a$ is real and not necessarily a natural number, but a reduction to this case can be easily made – Belgi Aug 24 '14 at 11:45
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As suggested by user_of_math, Stirling approximation of $n!$ can be very useful; it write $$n! \simeq \sqrt{2\pi n}\Big(\frac {n}{e}\Big)^n$$ Rewriting you inequality using logarithms leads to a comparison of $$\frac {1}{2}\log(2\pi)+(n+\frac {1}{2})\log(n)-n$$ and $$a \log(n)$$ So, we can find the minimum value of $n$ above which the inequality is verified : for example, for $a=10$, $n=15$; for $a=100$, $n=126$; for $a=1000$, $n=1165$;for $a=10000$, $n=11201$;for $a=100000$, $n=109432$.

Claude Leibovici
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