When is it true that $a^{n}<n!$?

Question

How can I find the smaller $n\in\mathbb{N}$, which makes the equation true: $$a^{n}<n!$$

For example:

If $a=2$ then $\longrightarrow 2^{n}<n!$ when $n\geq 4$

If $a=3$ then $\longrightarrow 3^{n}<n!$ when $n\geq 7$

If $a=4$ then $\longrightarrow 4^{n}<n!$ when $n\geq 9$

If $a=5$ then $\longrightarrow 5^{n}<n!$ when $n\geq 12$

$$a,n\in\mathbb{N}$$ $$a^{n}<n!$$when$$n\geq ?$$

Answer 1

Since $n!\approx n^ne^{-n}\sqrt {2\pi n}$ (Stirling), we have $a^n\approx n!$ when $a\approx \frac ne\,\sqrt[2n]{2\pi n}$. Hence $$n=\lceil ea\rceil$$ should be good enough. Note that this gives $n=6$ for $a=2$, $n=9$ for $a=3$, $n=11$ for $a=4$, $n=14$ for $a=5$, so is just mildly subobtimal.

Answer 2

I find a equation:

$$ a^{n}<n! \Longleftrightarrow n \geq (2a)+\lfloor\frac{(a-1)}{2}\rfloor$$