Prove that if $k \in \mathbb {N}$ and $a>1$, then $\lim\limits_{n\to\infty} \frac{n^k}{a^n}=0$ Prove that if $k \in \mathbb {N}$ and $a>1$, then $\lim\limits_{n\to\infty} \frac{n^k}{a^n}=0$
I have this, I want to know if what I did is correct.
Let be  $a^n>1>0$, then:
$1>\frac{1}{a^n}$
Then:
$|\frac{1}{a^n}|<1$
Aplying archimedean property,for $ε>0$ exists $n \in \mathbb {N}$ such that:
$nε>n^{k+1}$
Then:
$ε>\frac{n^{k+1}}{n}=n^k$ 
Let be $ε>0$, exists $N=max(1,ε)$ that if $n>N$
$|\frac{n^k}{a^n}-0|=|\frac{n^k}{a^n}|$
$=\frac{n^k}{a^n}$
$<ε (1)= ε$
Therefore the sequences converges to zero.
Thank, you.
 A: You must prove that for any $\varepsilon >0$ there exists $N\in\mathbb{N}$ such that for $n>N$ it happens that $$\frac{n^k}{a^n}<\varepsilon$$
This is quite hard to prove directly. To prove that 
$$\lim_{n\to\infty} \frac{n^k}{a^n}=0$$
We can use L'Hopital rule $k$ times and find that
$$\lim_{n\to\infty} \frac{n^k}{a^n}=\lim_{n\to\infty} \frac{kn^{k-1}}{a^n(\log a) }=\ldots=\lim_{n\to\infty} \frac{k!}{a^n(\log a)^k}=0$$
Hope this helps
A: Let's assume for a second the following is true $$\lim\limits_{n\rightarrow \infty}\frac{\ln{n}}{n}=0 \tag{1}$$
Now, let's look at the original problem from a different perspective, e.g.
$$\ln{\left(\frac{n^k}{a^n}\right)}=\ln{\left(n^k\right)}-\ln{\left(a^n\right)}=k\ln{\left(n\right)}-n\ln{\left(a\right)}$$
$$k\ln{\left(n\right)}-n\ln{\left(a\right)}<-k\ln{\left(n\right)} \Leftrightarrow 2k\ln{\left(n\right)} < n\ln{\left(a\right)} \Leftrightarrow \frac{\ln{n}}{n}<\frac{\ln{a}}{2k} \tag{2}$$
given $(1)$ is true, $(2)$ is true as well from some $n$ onwards. But then, from some $n$ onwards we have
$$\ln{\left(\frac{n^k}{a^n}\right)} < -k\ln{\left(n\right)} \Leftrightarrow 0<\frac{n^k}{a^n} < \frac{1}{e^{k\ln{n}}}=\frac{1}{n^k}$$
and the result follows from the squeeze theorem.

How do we prove $(1)$ though? Well, there are plenty of examples of this proof, for example this one.
