Cyclic subspace and one dimensional range 

*$\therefore$.. Let $T$ be a linear operator on a finite dimensional vector space $V$ with $\operatorname{dim} V$; $\geq 2$, and let $R(T)$ be the range of $T$. If $\operatorname{dim} R(T)=1$, prove that there exists a scalar $k$ such that $T^{2018}=k^{2017} T$
What I am thinking since range is one dimensional it is span of one vector $w$ so 
$T(v) = kw $ for some $k$ so 
$T^{2}(v) = T(kw) = kT(w)$ $= kw$ so for k =2 we are done.
Now I am thinking of using induction ? Am I right here?

We can write $Tx=f(x)w$ where $w$ is some non-zero vector and $f$ is  a linear map: $V \to \mathbb R$ (assuming that your vector space is over $\mathbb R$). Also $Tw=cw$ for some scalar $c$. This gives $T^{m}x=f(x)c^{m-1} w$ by induction . Hence  $T^{m}=c^{m-1}T$ for all positive integers $m$. Put $m=2018$. 
