Proof that the nth root of a rational s/t is irrational unless s and t are perfect nth powers with gcd(s,t)=1 A few months ago I asked for help with the above question (linked here Proof that the nth root of a rational s/t is irrational unless s and t are perfect nth powers) and I need a bit more help.

My lecturer has specifically said I need to use the uniqueness of prime factorisation, similar to that used in the proof that $\sqrt{2}$ is irrational via the uniqueness of prime factorisation seen below (taken from my project),

We can also think of a different proof for this as well. Taking $a$ and $b$ as before we end up with $2a^2 = b^2$. We shall now say that $a$ and $b$ have the following prime factor decomposition's,

$a = \alpha_1^{\beta_1}\cdot\alpha_2^{\beta_2}\cdot\alpha_3^{\beta_3}\cdot\dots\cdot\alpha_n^{\beta_n}$

and 

$b = \gamma_1^{\delta_1}\cdot\gamma_2^{\delta_2}\cdot\gamma_3^{\delta_3}\cdot\dots\cdot\gamma_n^{\delta_n}$

we see that when we have $2a^2 = b^2$ we can rewrite this as

$2\cdot\alpha_1^{2\beta_1}\cdot\alpha_2^{2\beta_2}\cdot\alpha_3^{2\beta_3}\cdot\dots\cdot\alpha_n^{2\beta_n} = \gamma_1^{2\delta_1}\cdot\gamma_2^{2\delta_2}\cdot\gamma_3^{2\delta_3}\cdot\dots\cdot\gamma_n^{2\delta_n}$

from this we can see that the left hand side must have an odd power two when $2$ and $a$ are multiplied together and the right hand side will have an even power (or no power at all). If these two expressions are equal then this is impossible due to the uniqueness of prime factorisation and we can once again conclude that $\sqrt{2}$ is irrational.

I either need to refine some of the answers given or redo the proof completely.

Any help greatly appreciated
Assume you have $\sqrt[n]{\frac{s}{t}}={a\over b}$, where s and t share no common primefactor. This leads to 

$$
t\cdot a^n=s\cdot b^n\space$$

Now since $s$ and $t$ are coprime, if you take any primefactor of $t$ it has to be a primefactor of $b$ and so occurs as a multiple of $n$ in the primefactorization of $t\cdot a^n$ and  hence also as a multiple of $n$ in the primefactorization of $t$. This shows that $t$ is a n-th power. 

The same argument can be applied to show that $s$ is a n-th power.
Assume you have $\sqrt[n]{\frac{s}{t}}={a\over b}$, where s and t share no common primefactor. This leads to 

$$
t\cdot a^n=s\cdot b^n\space$$

Now since $s$ and $t$ are coprime, if you take any primefactor of $t$ it has to be a primefactor of $b$ and so occurs as a multiple of $n$ in the primefactorization of $t\cdot a^n$ and  hence also as a multiple of $n$ in the primefactorization of $t$. This shows that $t$ is a n-th power. 

The same argument can be applied to show that $s$ is a n-th power.