The hierarchy of growth

1. Pick $q\in(L,1)$. Since $\dfrac{u_{n+1}}{u_n}\to L$, for $\varepsilon = q - L$, there is $N$ such that
$$\dfrac{u_{n+1}}{u_n}\leq L + \varepsilon = q, \, \, \forall n\ge N.$$
Hence
$$u_{n+1} \le qu_n, \, \, \forall n\ge N.$$
By induction, we get $u_n\le u_N q^{n-N}\to0$ as $n \to +\infty$.

2.
- For $u_n=\frac{n^k}{a^n}$: $\frac{u_{n+1}}{u_n}=\left(1+\frac1n\right)^k\frac1a\to\frac1a<1$.
- For $u_n=\frac{a^n}{n!}$: $\frac{u_{n+1}}{u_n}=\frac a{n+1}\to0<1$.
- For $u_n=\frac{n!}{n^n}$: $\frac{u_{n+1}}{u_n}=\frac{(n+1)n^n}{(n+1)^{n+1}}=\left(1+\frac1n\right)^{-n}\to\frac1e<1$.

Conclusion. For large $n$,
$$\ln n\ll n^k\ll a^n\ll n!\ll n^n.$$

Warning. If $L=1$ the test says nothing: $u_n=\frac1n$ tends to $0$, while $u_n=n$ diverges, though both have ratio $\to1$.