Vector space over $\mathbb{C}$ of all complex valued differentiable functions of a real variable $t$ basis? 
Let $V$ be the vector space over $\mathbb{C}$ consisting of all complex valued differentiable functions of a real variable $t$. Let $a_1,...,a_m$ be distinct complex numbers. The functions
$e^{a_1t},...e^{a_mt}$
are eigenvectors of the derivative, with distinct eigenvalues $a_1,...a_m$ and hence are linearly independent. Linear Algebra, by Serge Lang

I was wondering which basis generates this vector space. I think that $\dim V=1$ and a possible  basis could be  $e^t$.
Question:
Is $f(t)=e^t$ a basis of $V$?
Thanks in advance!
 A: No that's wrong. In fact what you quote is a way to prove that it is an infinite dimensional vector space.
In what you quote, it is explicitly said that whenever you pick distinct complex numbers $a_1,\dots,a_m$, the functions $e^{a_it}$ are linearly independent. So this means that your basis has at least $m$ vectors. As $m$ is arbitrary, this means that your vector space is infinite dimensional.
Edit: For curiosity, your quote shows something stronger. Your vector space doesn't have a countable basis.
A countable basis for an infinite dimensional vector space $V$ is a sequence of vectors $(v_n)_{n\in\mathbb{N}}$ such that:
1- For every vector $m\in\mathbb{N}$, the family $\{v_1,\dots,v_m\}$ is linearly independent.
2- For every vector $v\in V$, there is $m\in\mathbb{N}$ such that $v\in\text{span}\{v_1,\dots,v_m\}$.
For example, the vector space $\mathbb{C}[X]$ of all polynomials with complex coefficients has a countable basis, namely $(X^n)_{n\in\mathbb{N}}$.
But your vector space doesn't have a countable basis. That's because the uncountable family $\{e^{zt}\mid z\in\mathbb{C}\}$.
Nevertheless, the fact that every vector space has a basis is proved using Zorn's lemma.
