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?

Since
$$\lim_{n\to\infty}q_n=q<1,$$
therefore there exist such $k_0$ and $q<q'<1$ that
$$q_n<q', k\geq k_0.$$
Hence
$$\lim_{n\to\infty}n^kq_n^n\leq\lim_{n\to\infty}n^k(q')^n=0,$$
after root test:
$$(n^k(q')^n)^{\frac{1}{n}}=(n^{\frac{1}{n}})^k q'\to q' <1.$$
Moreover
$$\sum_{n=1}^{\infty}n^kq_n^n<\infty.$$
