I found this on Priestley's Complex Analysis in the Laplace transforms bit.
Suppose $f$ satisfies $f'(t)=f(kt)$ for $t>0$, where $0<k<1$ and $f(0)=1$. Prove that $$f(t)=\sum_{n=0}^{\infty}\frac{k^{n(n-1)/2}}{n!}t^n$$
Applying the Laplace transform directly to $f'(t)=f(kt)$ gives a functional equation but I'm unsure how to solve it. Any hints?
I do not know if the application of the Laplace transform would lead to a quicker solution, however, effectively, it is required to prove that $$f^{(n)}(0)=k^{\frac{n(n-1)}{2}}$$ In order to show this we can prove that $$f^{(n)}(t)=k^{\frac{n(n-1)}{2}}f(k^nt)$$ That it is true for $n=0$ and $n=1$ is easily verified. Now for $n\ge 1$ $$f^{(n+1)}(0)=k^{\frac{n(n-1)}{2}}\left(f(k^nt)\right)'=k^{\frac{n(n-1)}{2}+n}f(k^{n+1}t)=k^{\frac{n(n+1)}{2}}f(k^{n+1}t)$$ So the result holds by induction.