How to prove $n! > n^a$ for all $a\in \mathbb{R}$ (for sufficiently large $n$)?

Question

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 $(n+1)^a$:

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

Answer 1

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

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

Answer 2

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 $a=100000$, $n=109432$.