I think you need the spectral theorem (the fact that a normal operator is diagonalisable) for this. If you take an orthonormal basis consisting of eigenvectors of T, then the null space of T is spanned by the eigenvectors corresponding to the eigenvalue 0, and the image of T is spanned by the eigenvectors corresponding to nonzero eigenvalues. It's fairly easy to see that the same description works for the null space and image of any positive power of T.