$\{b^n\}\ll \{n!\}\ll n^n$

Fact 1. Let $b>0$, we have $\dis \limn \frac{b^n}{n!}=0$.

Proof. Let $x_n=b^n/n!$. Since $\limn x_{n+1}/x_n = \limn  \frac{b}{n+1}=0$, by ratio test, $\sum_{n=1}^\infty x_n$ converges, by divergence test, $\limn x_n=0$.$\qed$

Fact 2.  $\dis \limn \frac{n!}{n^n}=0$.

Proof. Observe that \[

\frac{n!}{n^n}=\brac{\frac{n}{n}\cdot \frac{n-1}{n}\cdot \frac{n-2}{n}\cdots \cdot \frac{2}{n}}\cdot \frac{1}{n}\leq \brac{1\cdot 1\cdots 1}\frac{1}{n}=\frac{1}{n},

\] so by Sandwich theorem, $\dis \limn \frac{n!}{n^n}=0$.$\qed$