Proving the limit of the sequence $1/n^r$ is zero where $r$ is a natural number I get this is a very easy proof but I'm trying to practice. Basically, I'm looking for feedback on this proof and insight on a possibly different way of doing this. Apologies in advance about the formatting.
My work:
By induction, consider $r=1$. This gives us the sequence $\frac{1}{n}$. By previous work, we know this sequence converges to $0$. Assuming $\frac{1}{n^k}$ is true for all $k$ in the natural numbers. I want to prove $\frac{1}{n^{k+1}}$ is true. This will establish that $\frac{1}{n^{r+1}}$ is also true. Let $ε$ be an arbitrary positive real number. Choose $N^{k+1}>\frac{1}{ε}$ where $N$ is also a natural number. Then whenever $n^{k+1}≥N^{k+1}$ we have: 
$|\frac{1}{n^{k+1}}-0|=\frac{1}{n^{k+1}}≤\frac{1}{N^{k+1}}<ε$. 
Thus, the limit of $\frac{1}{n^r}$ is $0$.

For all $n \in \mathbb{N}$
$$n \leq n^{r}$$
for all $r \in \mathbb{N}$ fixed. So
$$\frac{1}{n^{r}} \leq \frac{1}{n}$$
for all $r \in \mathbb{N}$ fixed.
If you prove that $\lim \frac{1}{n} = 0$, the result follows for all $r \in \mathbb{N}$.
