To show that $a^n \in O(n!)$

Question

$\forall n \in \mathbb{N} , a \in \mathbb{R} \wedge a > 0 $ $$a^n \in O(n!)$$

I need to show it. Probably I should use the Induction but I am not sure, anyone has tips for me please?

Answer 1

Try logarithms:

$\begin{align*} \ln a^n &= n \ln a \\ \ln n! &= \sum_{1 \le k \le n} \ln n \\ &\ge \frac{n}{2} \ln \frac{n}{n} \\ &= \frac{n}{2} \ln n + \frac{n \ln 2}{2} \end{align*}$

The estimate for $\ln n!$ is to take just the second half of the sum, and there the summands are at least $\ln (n / 2)$.