Is this proof that $e^z$ is transcendental correct? Lemma.  If $p(z)$ is a non-zero polynomial and $k\geq 1$ a natural number, then there exists a polynomial $q(z)$ of the same degree such that $\frac{d}{dz}\Big(p(z)\cdot e^{kz}\Big) = q(z)\cdot e^{kz}$.  (Indeed,
\begin{align}
\frac{d}{dz}\Big(p(z)\cdot e^{kz}\Big) &= p'(z)\cdot e^{kz} + kp(z)\cdot e^{kz} \\
&= e^{kz}\Big(p'(z) + kp(z)\Big),
\end{align}
and $p'(z) + kp(z)$ has the same degree as $p(z)$.)$\quad\square$
Recall that $f(z)$ is an algebraic function iff there exists a polynomial $F(z, y)\in\mathbb{Q}[X, Y]$ such that $F(z, f(z))\equiv 0.$
Assume that $e^z$ is algebraic and let
\begin{equation}
F(z, y) = \sum_{k=0}^n p_k(z)y^k = p_0(z) + \sum_{k=1}^n p_k(z)y^k,\quad p_n(z)\not\equiv 0,
\end{equation}
be a polynomial of minimal degree in $y$ such that $F(z, e^z) \equiv 0$.
Now, $p_0(z)$ cannot be the zero polynomial, or else $F(z, y) = y\cdot G(z, y)$, where $G$ is a polynomial of strictly lesser degree in $y$ than $F$ such that $G(z, e^z)\equiv 0$ (this because $e^z$ vanishes nowhere), contradicting the minimality of $deg_y(F)$.  So call $d:= deg(p_0\big(z)\big)$ and take $\frac{d^{d+1}}{dz^{d+1}}$ of both sides of the equation $F(z, e^z) \equiv 0$.  By linearity of derivative, the lemma, and the fact that the $(d+1)^{th}$ derivative of a degree $d$ polynomial is identically zero, this results in the equation
\begin{equation}
\sum_{k=1}^n q_k(z)e^{kz} \equiv 0,
\end{equation}
where $\deg(q_k) = \deg(p_k)$ for all $k\geq 1$.  We may factor $e^z$ out of this equation to obtain
\begin{equation}
e^z\cdot\sum_{k=0}^{n-1} q_{k+1}(z)e^{kz} \equiv 0,
\end{equation}
and since $e^z$ vanishes nowhere this equality implies that
\begin{equation}
\sum_{k=0}^{n-1} q_{k+1}(z)e^{kz} \equiv 0.
\end{equation}
But this implies that $\deg_y(F)$ was not minimal after all, contradiction.  So $e^z$ must be transcendental, ce qu'il fallait démontrer.  $\clubsuit$
 A: Yes, this is fine, although note that

*

*you never use this fact that $q(z)$ has the same degree as $p(z)$ so you don't need to mention it, and

*your definition of an algebraic function is slightly incorrect: $F$ should live in $\mathbb{C}[z, y]$. Fortunately this doesn't affect your argument, which proves that $e^z$ is not algebraic in this stronger sense. (Edit: this is not the definition used by Wikipedia or MathWorld, although it is used by PlanetMath. Personally I think Wikipedia and MathWorld are just wrong here. The Wikipedia article makes no use of the hypothesis that the coefficients are required to be in $\mathbb{Q}$ and everything it has to say applies to algebraic functions over $\mathbb{C}$. I suppose one can just specify the ground field but I still personally think "algebraic function," with no qualifications, should mean over $\mathbb{C}$.)

For fun, here is an alternative argument. We will just directly show that if $F$ is a nonzero polynomial then $F(z, e^z)$ is not identically zero. Suppose the monomial of largest degree that occurs in $F$ is $z^n e^{mz}$ (where largest is with respect to the lexicographic order where $m$ takes precedence over $n$; that is, we first consider the terms with $m$ maximal, then among those take the term with $n$ maximal). Then
$$\lim_{z \to \infty} \frac{F(z, e^z)}{z^n e^{mz}} \neq 0$$
(it must equal the coefficient of $z^n e^{mz}$ because $z^n e^{mz}$ dominates all other terms), so $F(z, e^z) \neq 0$. Here the limit to $\infty$ should be understood as happening along the positive real axis. However, this argument really requires that we're working over a subfield of $\mathbb{C}$, whereas the differentiation argument works over any field of characteristic zero.
