Show $\ker(\alpha)=\ker(\alpha)^2 \ \iff \ \ker(\alpha)\cap \mathrm{Im}(\alpha)=\{0\}$

Consider $v\in\ker \alpha^2.$ So $\alpha(\alpha(v))=0,$ that is, $\alpha(v)\in \ker\alpha.$ Moreorever $\alpha(v)\in\alpha(V).$ Since $\ker \alpha\cap \alpha(V)=\{0\}$ it must be $\alpha (v)=0,$ that is, $v\in \ker \alpha.$ This shows that $\ker\alpha^2\subset \ker \alpha.$

If $v$ is a nonzero vector in both $im(\alpha)$ and $ker(\alpha)$ then let $u$ be such that $\alpha(u)=v$. Then $\alpha(u)=v\not=0$ so $u\notin ker(\alpha)$, but $\alpha^{2}(u)=\alpha(v)=0$ so $u\in ker(\alpha^{2})$.

If $u\in ker(\alpha^{2})\backslash ker(\alpha)$ then $\alpha^{2}(u)=0$ so $\alpha(u)\in ker(\alpha)$ and $\alpha(u)\not=0$ so $\alpha(u)\in im(\alpha)\bigcap ker(\alpha)$ and is nonzero.