Prove $ \ \sqrt[5]{17} \notin \mathbb{Q}$ Question:

Prove that:
  $$ \ \sqrt[5]{17} \notin \mathbb{Q}.$$

My attempt:
Assume $ \ \sqrt[5]{17} \in \mathbb{Q}$. Then $ \sqrt[5]{17} = \dfrac{m}{n}$ where $ \ m,n \in \mathbb{Z}, n\neq 0$ and $ \ m,n$ have no common factors.
$ \ \sqrt[5]{17} = \dfrac{m}{n} \implies 17 = \dfrac{m^5}{n^5} \implies 17n^5 = m^5 \implies 17 | m^5 \implies 17|m \implies \exists a \in \mathbb{Z}$ s.t $ \ m = 17a$
Then, 
$ \ 17n^5 = m^5 \implies 17n^5 = 17^5. a^5 \implies n^5 = 17^4.a^5$
I am stuck here. $17^4$ is not prime. I need to show that $17 |n^5 \implies 17|n$
 A: In your proof if $n^5$ divided by $17^4$ then $n^5$ divided by $17$ and from here $n$ divided by $17$, which gives the contradiction.
I like the following way.
If $\sqrt[5]{17}=\frac{m}{n}$, where $m$ and $n$ are natural numbers then $17n^5=m^5$, which says $m$ divided by $17$.
Thus, $m=17m_1$, where $m_1$ is a natural number and from here we obtain:
$$n^5=17^4m_1^5,$$
which says that $n^5$ divided by $17$ 
and from here we obtain that $n$ divided by $17$ and $n=17n_1$, where $n_1$ is a natural number.
Thus, $17n_1^5=m_1^5$, where $m>m_1$.
Since we can repeat now this thing more and more,
we'll get an infinite series of natural numbers $\{m_i\}$ for which
$$m>m_1>m_2>...,$$
which is impossible.
We got a contradiction.
Thus, $\sqrt[5]{17}$ is an irrational number. 
A: If $p|x^n$ where $p$ is prime and $x,n$ are integers, then the prime factorization of $x^n$ has a $p^j$ term in it. The $n$th root of $x^n$ is an integer, so the $n$th root of $p^j$ must also be an integer (can you verify this?). Therefore $n|j$ and $j/n \geq 1$ since $p$ does not divide $1$.
