Why is $e^g$ transcendental over $\mathbb{C}(z)$ for $g$ rational? In the 1972 paper Integration in Finite Terms by M Rosenlicht he says:

we note that if $g(z)$ is is a non-constant rational function of the complex variable $z$ then $e^g$ is not algebraic over $\mathbb{C}(z).$ [...] it can be shown algebraically by looking at the irreducible equation over $\mathbb{C}(z)$ that $e^g$ would otherwise satisfy,
$e^{ng}+a_1e^{(n-1)g}+\dotsb+a_n=0$
Then differentiating this to get
$ng'e^{ng}+(a_1'+(n-1)a_1g')e^{(n-1)g}+\dotsb+a_n'=0,$
Which must be proportional to the first equation, so that $ng'=a_n'/a_n$, then noting that $a_n'/a_n$ is either zero or a sum of fractions with constant numerators and linear denominators, whereas $ng'$ can have no linear denominator, so that $g'=0,$ a contradiction.

See screencap.

I can't understand this proof. Why should the two equations be proportional? If it were a polynomial $P(y)$ with constant coefficients evaluated at $y=e^g$, then differentiating would give $P'(e^g)e^gg'$ which would still not make the two equations proportional. Anyway the polynomial doesn't have constant coefficients so that's not what's going on here.
Or if you have your own proof of the transcendence of the exponential of a rational function (using differential algebra, not analytic methods) I'd accept that instead.
 A: Write $\alpha=e^g$.
If $\alpha$ satisfies $P(\alpha)=0$ then $P$ is divisible by $\alpha$'s minimal polynomial. If further $P$ is irreducible, then $P$ must be the minimal polynomial. If $Q(\alpha)=0$ for another polynomial $Q$ of the same degree, then $Q$ must be proportional to $P$, since otherwise $q_nP-p_nQ$ (where  $p_n$ and $q_n$ are the leading coefficients of $P$ and $Q$ respectively) would be a nonzero polynomial with $\alpha$ as a root having degree smaller than $\alpha$'s minimal polynomial, a contradiction.
If $P(t)=t^n+\cdots+a_n$ and $Q(t)=ng't^n+\cdots+a_n'$ are proportional, then the constant of proportionality must be $ng'$ (compare leading coefficients), which means $ng'$ times $a_n$ is $a_n'$.
A: The point is that $P(\zeta) = \zeta^n + a_1 \zeta^{n-1} + \ldots + a_n $ has minimal degree among polynomials over $\mathbb C(z)$ that vanish at $\zeta = e^g$.  But taking the derivative of $P(e^{g})$ we get another polynomial 
$Q(\zeta) = n g' \zeta^n + (a_1' + (n-1) g' a_1) \zeta^{n-1} + \ldots + a_n'$ 
of degree $ \le n$ over $\mathbb C(z)$ that vanishes at $e^{g}$.  If this were not proportional to $P(\zeta)$, we could take a linear combination of them to get a nontrivial polynomial of lower degree that vanishes at $e^{g}$, contradicting minimality.
