If $p>0$, then $ \lim_{n\to\infty}\frac{1}{n^p}=0$ using squeeze theorem for sequences. 
If $p>0$, then $ \lim_{n\to\infty}\frac{1}{n^p}=0\;.$

Rudin suggests in his Principle of Mathematical Analysis to take  $$n> (\frac{1}{\epsilon})^\frac{1}{p}$$ using the Archimedean property of the real number system.
This is under the assumption that we will compute the limit of the sequence based on the fact: If $\ 0 \leq x_n \leq s_n$ for $\ n \geq N$, where N is some fixed number, and if $\ s_n \rightarrow 0$, then $\ x_n \rightarrow 0.$
I don't really understand this proof, but I could try it a different way: 
Letting $\ x_n = \frac{1}{n^p}$ and taking $\ s_n$ to be $\ \frac{1}{n}$, then we know that $\ x_n \leq s_n $ because $\ p>0 $. But since $ s_n = \frac{1}{n}$ goes to $0$ as $n$ approaches infinity, we know from $0 \leq x_n \leq s_n$ that $ x_n = \frac{1}{n^p}$ will also go to $0$ as $n$ approaches infinity.
Is this a valid approach?

Is Rudin serious? Do you acutally prove 

If $p>0$, then $ \lim_{n\to\infty}\frac{1}{n^p}=0\;.$

using the squeeze theorem?
Well, he might have been teasing the student a bit with that one. The most elementary/first principle thing we learn is that
$\tag 1 \displaystyle \lim_{n\to\infty}\frac{1}{n}=0$
and that it follows from the Archimedean property. So one 'heartbeat away' we demonstrate that
$\tag 2 \displaystyle \lim_{n\to\infty}\frac{1}{n^p}=0$
once again using the Archimedean property.
So how does Rudin justify bringing out the squeeze theorem here?
Well, just set $x_n = \frac{1}{n^p}$ and $s_n = \frac{1}{n^p}$ (you've seen in Rudin's proof that that $s_n$ converges to $0$).
I think Rudin would have made a great lawyer!
