Simple proof that $z\mapsto \exp(i\langle \omega,z\rangle)$ are linearly independent I am interested in proving that the family of functions
$$\{f_{\omega}: \mathbb{C}^n\rightarrow\mathbb{C}, f_\omega(z) = \exp(i\langle \omega, z \rangle): \omega \in \mathbb{C}^n\},$$
where $\langle \cdot,\cdot\rangle$ is the usual hermitian dot product, is $\mathbb{C}$-linearly independent.
In the case $n=1$ an expeditive argument consists in remarking that these functions are eigenfunctions with distinct eigenvalues of the complex derivation operator.
Is there a somewhat similar argument, or a simple way to prove the result in dimension $n$ ?
 A: a) Changing your notation slightly, we must prove that if  $L_i:\mathbb C^n\to \mathbb C $ are  a finite set of mutually distinct linear forms, then  the entire functions  $\exp (L_i):\mathbb C^n\to \mathbb C$ are linearly independant. 
b) If we could find $u\in \mathbb C^n$ such that the numbers  $L_i(u)\in \mathbb C$ are distinct we would have solved our problem:
Indeed any linear relation $\sum c_i\exp (L_i(z))=0$ between the  $\exp (L_i)$ would imply (by substituting $\zeta\cdot u$ for $z$) that $\sum c_i\exp [L_i(u)\cdot\zeta]=0$ for all $\zeta\in \mathbb C$ and this in turn  implies that all the $c_i=0$ by the $n=1$ version of your question.
Thus the $\exp (L_i)$'s would indeed have been proved linearly independent. 
c) We are thus reduced to the problem in pure linear algebra of showing that given  finitely many linear forms $L_i$ on $\mathbb C^n $ there exists a vector $u\in \mathbb C^n $ on which they take different values: $$L_i(u)\neq L_j(u) \:\text {for}\: i\neq j.$$  But this is easy : take any $u\in \mathbb C^n $ outside of the union of the finitely many hyperplanes $\ker (L_i-L_j)$ !
A: Yes, you just need make make a slightly elaborate argument.
The function $E_\omega \in C(\mathbb{C}^n, \mathbb{C})$ defined by $z \mapsto \exp(i \langle \omega, z \rangle)$ is an eigenvector for each of the partial Derivative operators $\partial/\partial z_j$, $j = 1, \dots n$, and the corresponding eigenvalues are $\omega_j$.
Now assume that
$$ \lambda_1 E_{\omega^1} + \dots + \lambda_N E_{\omega^N} \equiv 0.$$
Inserting $z = (z_1, 0, \dots, 0)$, this becomes a statement about the one-dimensional case, where you can apply your argument. In particular, you conclude that some of the $\omega^k$ have the same first entry or all $\lambda_k$ are zero. This way you reduce the equation above to a number of equations of the same kind, but with smaller $N$ and the property that all $\omega^k$ have the same first entry.
Proceeding this way for $j= 1, \dots, n$, you finally get that each $\lambda_k$ is zero, unless some of the $\omega^k$ coincide.
