Prove $\lim_{n \to \infty}\frac{n^k}{a^n}=0 $ , if $a>1$ and $ k \in \mathbb{N}^+$ I was stuck in a basic real analysis homework, here is the situation:
The homework is:

Use the Stolz-Cesàro theorem to prove  $\lim_{n \to \infty}\frac{n^k}{a^n}=0  $, if $a>1$ and $k \in \mathbb{N}^+$.

what I have tried:
Let $x_n=n^k$ and $y_n=a^n$,
then $$\lim_{n\to\infty}y_n=+\infty \thinspace and \thinspace \{y_n\} \text{ is a strickly increasing monotone series}$$
based on this, I am trying to use the Stolz-Cesaro theorem, so I try to calculate the limit below: $$\lim_{n\to\infty}\frac{x_{n+1}-x_{n}}{y_{n+1}-y_n} \thicksim \frac{(n+1)^k-n^k}{a^{n+1}-a^n}$$
in this step, I try use binomial formula because the $n^k$ can be cancel:
$$\lim_{n\to\infty}\frac{x_{n+1}-x_{n}}{y_{n+1}-y_n} \thicksim \frac{(n+1)^k-n^k}{a^{n+1}-a^n}\thicksim\frac{(C_k^0n^0+C_k^1n^1+\cdots+C_k^{k-1}n^{k-1}+C_k^kn^k)-n^k}{a^{n}(a-1)}\thicksim???$$
then I stuck here.
I tried to find an answer online, which says that

...use the Stolz-Cesaro theorem k times, then you can reduce the power of n  to 0 th in the numerator while the power of a in the denominator will reduce to n-k then you will conclude a limit which is 0, then the question is solved.

But I am confused with the answer I found above. Can you give me a better solution or clue?
I know something I had learned in Calculus may solve it : Use L'Hospital's rule k times to solve this question.
But I haven't learn how to prove L'Hospital's rule in real analysis, I even haven't learn the definition of limit, continuity, derivative of functions in REAL ANALYSIS. So I think better not to use them now.
Can you help me? Thanks!!!
 A: Usually when you come across a statement of the form "use this k times..." there is probably an induction hidden in this statement, so we will try to prove this with induction on $k$ the power of $n$.

*

*Induction base: when $k=0$ the the desire limit is $\lim_{n\to\infty}\frac{1}{a^n}$. Notice that
$$
\frac{1-1}{a^{n+1}-a^n}=\frac{0}{a^n({a-1})}=0\xrightarrow{n\to\infty}{}0
$$
So by Stolz-Cesaro theorem $\lim_{n\to\infty}\frac{1}{a^n}=0$.

*Induction step: Now assume that $\lim_{n\to\infty}\frac{n^m}{a^n}=0$ for all $0\leq m<k$ and we will prove that $\lim_{n\to\infty}\frac{n^{k}}{a^n}=0$. As you observed

$$
\frac{(n+1)^{k}-n^{k}}{a^{n+1}-a^n}=\frac{\sum_{m=0}^{k}\binom{k}{m}n^m-n^{k}}{a^n(a-1)}=\frac{\sum_{m=0}^{k-1}\binom{k}{m}n^m}{a^n(a-1)}=\frac{1}{a-1}\cdot\sum_{m=0}^{k-1}\binom{k}{m}\frac{n^m}{a^n}
$$
From our induction hypothesis it follows that $\lim_{n\to\infty}\frac{n^m}{a^n}=0$ for $0\leq m<k$, so by limit arithmetic we will get
$$
\lim_{n\to\infty}\frac{1}{a-1}\cdot\sum_{m=0}^{k-1}\binom{k}{m}\frac{n^m}{a^n}=\frac{1}{a-1}\sum_{m=0}^{k-1}\binom{k}{m}\lim_{n\to\infty}\frac{n^m}{a^n}=\frac{1}{a-1}\sum_{m=0}^{k-1}\binom{k}{m}\cdot0=0
$$
So by Stolz-Cesaro we will conclude that $\lim_{n\to\infty}\frac{n^k}{a^n}=0$ as desired.
A: A direct proof is possible. Write $a = 1 + h$ with $h > 0$. Then use the binomial theorem to get that for $n \geq k + 1$,
$$(1 + h)^n = \sum_{i = 0}^{n}\binom{n}{i}h^i \geq \binom{n}{k + 1}h^{k + 1} \geq \frac{(n - k)^{k + 1}}{(k + 1)!}h^{k + 1}.$$
Since $\frac{n^k}{(n - k)^{k + 1}} \to 0$ as $n \to \infty$, the proof is finished.
