Let $E$ be an elliptic curve over an algebraically closed field $k$ of positive characteristic $p$. Recall that $[p]: E \rightarrow E$ is always an inseparable isogeny. Therefore, by the above, it factors through $F: E \rightarrow E^p$. Moreover $E$ is supersingular iff $E[p](k) = 0$ iff the Frobenius isogeny $F: E \rightarrow E^p$ [p]$is purely inseparableits , iff the dual isogeny (called "Verschiebung") to Frobenius$V: E^p \rightarrow E$(the "Verschiebung") is also inseparable. This But again, this means that the dual isogeny$V$factors through the Frobenius isogeny for$E^p$-- i.e.,$E^p \rightarrow E^{p^2}$-- and since both have degree$p$this means that$E$is isomorphic to$E^{p^2}$. Thus on$j$-invaraiants we have$j(E)^{p^2} = j(E)$, done. 1 (Note: the following argument uses the fact that an isogeny of elliptic curves is inseparable iff it factors through the Frobenius isogeny. This is a result in Silverman's book, for instance.) An elliptic curve$E$over an algebraically closed field$k$is supersingular iff $E[p](k) = 0$ iff the Frobenius isogeny$F: E \rightarrow E^p$is purely inseparable its dual isogeny (called "Verschiebung")$V: E^p \rightarrow E$is inseparable. This means that the dual isogeny$V$factors through the Frobenius isogeny for$E^p$-- i.e.,$E^p \rightarrow E^{p^2}$-- and since both have degree$p$this means that$E$is isomorphic to$E^{p^2}$. Thus on$j$-invaraiants we have$j(E)^{p^2} = j(E)\$, done.