The solution of this can go as follows.
THEOREM
If $(D-a)^n y=0$ then $y=\left(c_1+c_2 x+\cdots +c_n x^{n-1} \right) e^{ax}$
PROOF
We begin by the indentity, where $u$ is a function of $x$:
$\left(D-a \right)(e^{ax} u)=e^{ax} Du$
We show by induction that
$\left(D-a \right)^n(e^{ax} u)=e^{ax} D^nu$
It is true for $n=1$. We show the inductive step:
$(D-a)^n (e^{ax} u)=e^{ax} D^n u$
$(D-a)^{n+1} (e^{ax} u)=(D-u)e^{ax} D^n u$
$(D-a)^{n+1} (e^{ax} u)=ae^{ax}D^n u +e^{ax} D^{n+1} u-ae^{ax}D^n u$
so by the principle of induction
$(D-a)^n (e^{ax} u)=e^{ax} D^n u \text{ ; for all } n\in \Bbb N$
Consider then the equation
$(D-a)^n y = 0$
Then set $y=e^{ax} u$. We get that
$(D-a)^n (e^{ax}u) =e^{ax} D^n u= 0$
Then $D^n u =0$ from the theorem, so if
$u=\sum_{k=1}^n c_k x^{k-1}$ then $D^n u=0$, and since it has $n$ arbitrary cosntants, it is a general solution. Thus
$y= e^{ax} \sum_{k=1}^n c_k x^{k-1}$
Note that each individual term is also a solution.
A more general result is:
THEOREM (Operator shift formula)
Let $\phi(D) = a_0D^n+a_1 D^{n-1}+\cdots+a_{n-1}D+a_n$ Then
$\phi(D) \{e^{mx} F\} =e^{mx} \phi(D+m)\{ F\}$
PROOF
Firstly, $ \tag 1 D \{ e^{mx} F\} = e^{mx} (D+m) F$
Direct calculation
$\frac{d}{{dx}}\left( {{e^{mx}}F} \right) = {e^{mx}}\frac{d}{{dx}}F + {e^{mx}}mF = {e^{mx}}\left( {D + m} \right)F$
Secondly,
$\tag 2 {D^n}\left\{ {{e^{mx}}F} \right\} = {e^{mx}}{\left( {D + m} \right)^n}F$
By Lebniz formula
$ D^n \{e^{mx} F \} = \sum_{k=0}^n {n \choose k} D^k \{e^{mx} \} D^{n-k} \{F \}$
$ D^n \{e^{mx} F \} = \sum_{k=0}^n {n \choose k} m^k e^{mx} D^{n-k} \{F \}={e^{mx}}{\left( {D + m} \right)^n}F$
Finally, since $\phi(D)$ is linear, viz:
$\eqalign{ & \phi \left( D \right)\left\{ {F + G} \right\} = \phi \left( D \right)\left\{ F \right\} + \phi \left( D \right)\left\{ G \right\} \cr & \phi \left( D \right)\left\{ {k \cdot F} \right\} = k \cdot \phi \left( D \right)\left\{ F \right\} \cr} $
We get
$\phi \left( D \right)\left\{ {{e^{mx}}F} \right\} = {e^{mx}}\phi \left( {D + m} \right)\left\{ F \right\}$