sequences and series - Prove that if $k in mathbb {N}$ and $a>1$, then $limlimits_{ntoinfty} frac{n^k}{a^n}=0$

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.