Laplace transform exercise 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?
 A: 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.
A: Let $f(t)=\sum\limits_{n=0}^\infty a_nt^n$ ,
Then $\sum\limits_{n=0}^\infty na_nt^{n-1}=\sum\limits_{n=0}^\infty a_n(kt)^n$
$\sum\limits_{n=1}^\infty na_nt^{n-1}-\sum\limits_{n=0}^\infty k^na_nt^n=0$
$\sum\limits_{n=1}^\infty na_nt^{n-1}-\sum\limits_{n=1}^\infty k^{n-1}a_{n-1}t^{n-1}=0$
$\sum\limits_{n=1}^\infty(na_n-k^{n-1}a_{n-1})t^{n-1}=0$
$\therefore na_n-k^{n-1}a_{n-1}=0$
$a_n=\dfrac{k^{n-1}a_{n-1}}{n}$
$\therefore\begin{cases}a_0=a_0\\a_n=\dfrac{(k^0k^1k^2......k^{n-1})a_0}{1\times2\times3\times......n}\forall n\in\mathbb{N}\end{cases}$
$\begin{cases}a_0=a_0\\a_n=\dfrac{k^{0+1+2+......+(n-1)}a_0}{n!}\forall n\in\mathbb{N}\end{cases}$
$\begin{cases}a_0=a_0\\a_n=\dfrac{k^{\frac{n(n-1)}{2}}a_0}{n!}\forall n\in\mathbb{N}\end{cases}$
$a_n=\dfrac{k^{\frac{n(n-1)}{2}}a_0}{n!}\forall n\in\mathbb{Z}^*$
$\therefore f(t)=C\sum\limits_{n=0}^\infty\dfrac{k^{\frac{n(n-1)}{2}}t^n}{n!}$
$f(0)=1$ :
$C=1$
$\therefore f(t)=\sum\limits_{n=0}^\infty\dfrac{k^{\frac{n(n-1)}{2}}t^n}{n!}$
