If the rank$ (T)=1$ and Im $(T) \cap$ Ker$(T)$ is zero, show $T$ is diagnolizable. The full question

Let $T: V\rightarrow V$ be a linear operator. If the rank $(T)=1$ and Im$(T) \cap$ Ker$(T)=0$, show $T$ is diagonalizable.

Alright so I've been trying my hand at this the past few days now and I've got kind of an idea, but I'm still not sure how to prove this. 
What we know: 

If we're dealing with a dim$(V) = n$, we know that dim$(Ker(T)) = n-1$ since the dim(Im$(T))=1$. What this means is that we have a basis of eigenvectors for our null space s.t that we have $n-1$ corresponding eigenvalues; they all equal zero. In order to show that $T$ is diagnolizable, we have to find a basis for Im$(T)$ s.t it only contains one eigenvector and a corresponding non-zero eigenvalue independent of our nullspace.

So I kind of get it, but I have no idea how to properly state/prove this. If anyone could give me a hand, I'd love the help. It's been driving me nuts!

Since $\operatorname{rank}(T)=1$ then by the rank-nullity theorem we have
$$\dim\ker(T)=n-1$$
and since 
$$\operatorname{Im}(T)\cap \ker(T)=\{0\}$$
then
$$V=\operatorname{Im}(T)\oplus \ker(T)$$
now we take a basis of $V$ adapted to the previous decomposition and we see that the matrix of $T$ in this basis take the form
$$\operatorname{diag}(\alpha,\underbrace{0\ldots,0}_{n-1\;\text{times}})$$
 where $\alpha=\operatorname{tr}(T)$ hence $T$ is diagonalizable.
