For $0I have problems with computing the following limit:

Given a sequence $0<q_n<1$ such that $\lim\limits_{n\to\infty} q_n =q < 1$, prove that for a fixed $k \in \mathbb N$, $\lim\limits_{n\to\infty} n^k q_n^n= 0$.

I know how to prove this, but I can't do it without using L'Hôpital's Rule. Does someone have an elementary proof?

Note that $$\frac{(n+1)^k}{n^k}=1+kn^{-1}+{k\choose2}n^{-2}+\ldots+n^{-k}\to1$$
as $n\to\infty$, hence for any $s$ with $1<s<\frac1q$ (possible because $q<1$) we can find $a$ with $n^k<a\cdot s^n$.
Select $r$ with $q<r<\frac1s$ (possible because $s<\frac1q$). Then 
For almost all $n$, we have $q_n<r$, hence
$$n^kq_n^n<n^kr^n<a(rs)^n.$$
Since $0<rs<1$, the claim follows.
