I’d like to show that the space of $C^1$ solutions $f:\mathbb{R} \rightarrow \mathbb{C}$ of the delay differential equation $f’(x)=f(x+1)$ is infinite-dimensional.
I think this can be done by using the fact that there exists an infinite number of solutions to the equation $e^z-z=0$, where $z \in \mathbb{C}$, which I’ve already shown… but I haven’t managed to find the good argument so far.
Edit: I guess the idea must be something like assuming the solution is $ f(x)= {\rm e}^{\lambda x } $ and substituting in the equation to get $ \lambda {\rm e}^{\lambda x } = {\rm e}^{\lambda (x+1) }$, that is $\lambda = {\rm e}^{\lambda } $, which as I said has infinitely many complex solutions. But I’m not sure this is all there is to it…
It suffices to find infinitely many solutions that are linearly independent.
As you point out, each root $\lambda$ of the equation $$e^{\lambda}=\lambda\tag{*}$$ gives a solution $f(x)=e^{\lambda x}$; you need to show that these are linearly independent.
One way to do this is to consider a purported dependence relation: $$0=\sum_{j\in J}{a_je^{\lambda_j x}}$$ where (1) $J$ is finite, (2) each $\lambda_j$ solves (*), and (3) there is no $j$ for which $a_j=0$.
Pick the $\lambda_j$ with largest real part and divide by $e^{\lambda_j x}$. Now take $x\to\infty$. What happens?