Does $\operatorname{im}(\alpha^2)=\operatorname{im}(\alpha)$ finite-dimensional imply $V=\ker(\alpha)\oplus \operatorname{im}(\alpha)$? 
Prove or disprove: Let $\alpha$ be a linear operator on $V$. If $\operatorname{im}(\alpha)=\operatorname{im}(\alpha^2)$ and $\operatorname{im}(\alpha)$ is finite-dimensional, then $V=\ker(\alpha)\oplus \operatorname{im}(\alpha)$.

The usual counterexample uses the left shift operator, whose image is, of course, infinite-dimensional. The claim is true if we require $V$ to be finite-dimensional (using the rank-nullity theorem). It is also true if we require the stronger condition that $\alpha^2=\alpha$. Can anyone help?

Yes, this is true. $\def\im{\operatorname{im}}\def\a{\alpha}$
The key fact is that $\a$ restriced to $\im(\a)$ is an isomorphism: $\a:\im(\a)\to \im(\a)$ is surjective because $\im(\a)=\im(\a^2)$, and a surjective map of finite-dimensional vector spaces must be an isomorphism by rank-nullity.
Now we can dispatch the problem quickly. First, $\im(\a)\cap \ker(\a)=\{0\}$: let $v\in V$ belong to both, then $\a(v)=0$ which implies $v=0$ as $\a$ is an isomorphism on $\im(\a)$.
We also need to show every element in $V$ has a decomposition as the sum of an element in the kernel and an element in the image. Again let $v\in V$, and let $w\in\im(\a)$ be the unique element so that $\a(w)=\a(v)$ which exists because $\a$ is an isomorphism on $\im(\a)$. Then $v-w$ and $w$ are such a decomposition: $\a(v-w)=\a(v)-\a(w)=0$, so $v-w\in\ker(\a)$, and $w\in\im(\a)$. Thus we've shown the necessary conditions for $V=\ker\a\oplus\im\a$.
