Fermat's Curve is not rational (Perrin's "Algebraic Geometry - An Introduction") Below is the proof that curve given by $x^n+y^n=1$ for $n\geq 3$, over field of characteristic that does not divide $n$, has no rational parametrization, from Intrduction of Perrin's "Algebraic Geometry - An Introduction". I don't understand the final step.

Assume given a parametrization $x=p(t)/r(t)$, $y=q(t)/r(t)$ such that $p$, $q$, $r\in k\left[t\right]$ have no common factor. We would then have $p^n+q^n-r^n=0$, and hence $p$, $q$, $r$ are mutualy coprime. On differentiating this equation we get
  $$
p^{n-1}p'+q^{n-1}q'-r^{n-1}r'=0
$$
  We may suppose that the degree of $p$ is at least as large as the degrees of $q$ and $r$. After multiplying by $r$ we get
  $$
p^{n-1}(rp'-pr')=q^{n-1}(qr'-rq')
$$
  and since $p$ and $q$ are coprime, $p^{n-1}$ divides $qr'-rq'$, which, since $n\geq 3$, is impossible for degree reasons.

... well, unless $qr'-rq'=0$. Is there an obvious reason why it cannot be the case here, and the proof is complete? 
 A: Note that $r$ is non-zero and consider $$\left(\cfrac qr\right)'=\cfrac {(q'r-qr')}{r^2}$$ which would be zero in the case you cite, so that we would have $q(t)=kr(t)$
The possibility $r(t)=1$ (otherwise $r(t)$ is a common factor) has to be excluded to make the question tight. This may be implicit in the definition of a rational parametrisation. For example, if at least one of $p,q,r$ had to be non-constant, it would be impossible to have $r=1$.
A: Is there a blatantly obvious reason? No.
However, in this case you would have $p'/p = q'/q = r'/r$, implying either that every prime factor of $p$ must also be a prime factor of $q$ and $r$ (by looking at the poles, or if you prefer the factorization of the denominator), or that $p' = q' = r' = 0$, and thus $p,q,r$ are polynomials in $x^c$ where $c$ is the characteristic of the field. In this latter case, you can use the method of infinite descent, by substituting $x^c \to x$.
(there may be a hole in my proof regarding looking at the poles when the characteristic is positive)
