Are the functions $e^{z^n}$ all algebraically independent? I'm currently writing about Riemann surfaces and the algebraic dependence of any two meromorphic functions on a compact surface. I'm trying to think of an example of how this result fails for a noncompact surface, and my hunch is that the family $$ e^{z^n}, n = 1,2, ... $$ gives an infinite family of algebraically-independent functions over $\mathbb{C}$, equivalently that all functions $e^{P(z)}$ are linearly independent over $\mathbb{C}$ for $P$ positive integer polynomials. However I can't seem to get a proof out.
Am I being thick, and is there an obvious proof, or is it obviously untrue? If it's untrue, can anyone furnish an example of a family, preferably infinite, of algebraically-independent (over $\mathbb{C}$) meromorphic functions on the plane?
Many thanks in advance!
 A: Let $f_n(z)=e^{z^n}$. Assume $P(f_1,\ldots,f_n)=0$ with $P\in\Bbb C[X_1,\ldots, X_n]$.
Each monimial $a_{i_1,\ldots, i_n}X_1^{i_1}\cdots X_n^{i_n}$ of $P$ contributes $a_{i_1,\ldots, i_n}e^{i_1z+i_2z^2+\ldots+i_nz^n}$. As the polynomials $i_1X+i_2X^2+\ldots+i_nX^n\in\Bbb Z[X]$ are pairwise distinct and have no constant term, we know (see the claim below) that the $e^{i_1z+i_2z^2+\ldots+i_nz^n}$ are $\Bbb C$-linearly independent, hence all $a_{i_1,\ldots, i_n}$ are $=0$.

Here's the missing link:
Claim. The family  $\left\{t\mapsto e^{tf(t)}\right\}_{f\in\Bbb R[X]}$ of functions $\Bbb R\to\Bbb R$ is $\Bbb R$-linearly independent.
Proof. We can define a total order on $\Bbb R[X]$ by letting $$f\prec g\iff \exists x_0\in\Bbb R\colon \forall x>x_0\colon f(x)<g(x).$$ Equivalently, the order is determined by the sign of the leading coefficient of $f-g$.
Since non-zero polynomials are not asymptotically zero, we have
$$\tag1 f\prec g\implies \exists r>0\colon\exists x_0\in\Bbb R\colon\forall x>x_0\colon f(x)<g(x)-r$$
Assume the given family of functions is not independent and let 
$$\tag2 \sum_{i=1}^n \alpha_ie^{tf_i(t)}\equiv 0$$
be a linear dependence relation with minimal $n$. Then $n\ge 1$ and all $\alpha_i$ are non-zero. We may assume wlog. that $f_1\prec f_2\prec\ldots \prec f_n$. If we multiply $(2)$ with $e^{-tf_n(t)}$, we obtain
$$-a_n=\sum_{i=1}^{n-1}a_ie^{t(f_i(t)-f_n(t))} $$ 
Using $(1)$ we have $f_i(t)-f_n(t)<-r<0$ for $t\gg 0$, hence $e^{t(f_i(t)-f_n(t)}\to 0$ as $t\to +\infty$. We conclude $a_n=0$ contrary to our assumption. $\square$
Corollary. The family  $\left\{z\mapsto e^{zf(z)}\right\}_{f\in\Bbb R[X]}$ of functions $\Bbb C\to\Bbb C$ is $\Bbb C$-linearly independent. $\square$
