Then $F_a$ divides $F_{an}$, $n \in \Bbb{N}$ where $F_i$ denotes the $i$-th Fibonacci number. 
Let $a$ be a fixed natural number. Then $F_a$ divides $F_{an}$, $n \in \Bbb{N}$ where $F_i$ denotes the $i$-th Fibonacci number.

For $n=1$ the result is true. Let it be true for $n=k$.
$$\begin{align}F_{ka+a} &= F_{ka+(a-1)} + F_{ka+(a-2)} \\
&= 2F_{ka+(a-2)} + F_{ka+(a-3)}\\
&= 3F_{ka+(a-3)} + 2F_{ka+(a-4)}\\
&= 5F_{ka+(a-4)} + 3F_{ka+(a-3)} = ...\\
&= F_aF_{ka+1} + F_{a-1} F_{ka}\end{align} $$
Thus $F_a$ divides $F_{(k+1)a}$. Hence the result is true by induction.
Is the proof correct?

To avoid the dots in your proof, show by (strong) induction with respect to $n$, that a more general identity holds:
$$F_{n+m+1}=F_{m+1}F_{n+1}+F_{m}F_{n}.$$
Then apply it for $m=a-1$ and $n=ka$, to find directly what you need
$$F_{ka+a} = F_aF_{ka+1} + F_{a-1} F_{ka}.$$
For the inductive step:
$$\begin{align}
F_{n+m+2}&=F_{n+m}+F_{n+m+1}\\
&=(F_{m+1}F_{n}+F_{m}F_{n-1})+(F_{m+1}F_{n+1}+F_{m}F_{n})\\
&=F_{m+1}(F_n+F_{n+1})+F_m(F_{n-1}+F_n)\\
&=F_{m+1}F_{n+2}+F_mF_{n+1}
\end{align}$$
