real analysis - If a polynomial p(x) is divisible by (x-a)^n where nϵℕ and n≥2, then p(x) is divisible by (x-a)^(n-1)

I have no idea how to begin, I do know that if $p(x)$ is divisible by $(x-a)^n$ then we should have $p(x)=(x-a)^n*q(x)+r(x)$ where $r(x)=0$. And this seems like something I would want to try induction on but it seems like a reverse type of induction instead of proving that if $n$ then $n+1$, I'm doing if $n$ then $n-1$ which I don't really know how to approach. I could try contrapositive induction? If $p(x)$ is not divisible by $(x-a)^{n-1}$ where $n≥2$ then $p(x)$ is not divisible by $(x-a)^{n}$. I would rewrite it as $p(x)$ not divisible by $(x-a)^n$ where $n≥1$ then $p(x)$ not divisible by $(x-a)^{n+1}$. But how would I even write my induction steps out... I appreciate any insight and help given

Answer

Asserting that $p(x)$ is divisible by $(x-a)^n$ means that there is a polynomial $q(x)$ such that $p(x)=(x-a)^nq(x)$. But then $p(x)=(x-a)^{n-1}\bigl((x-a)q(x)\bigr)$.