Although I know the properties of linear transformations and the concept of rank, I could not prove this. Can you help me?
I chose an element from the kernel and said this element b. I saw that this element is a rangein element L (b) = sb for some numbers.
Remember that for a finite dimensional vector space $V$ we have $dim(V)=dim(ImF)+dim(kerF)$, where $F$ Is a linear map.
In your case $dim(ImF)=dimV-dim(KerF)=n-(n-1)=1\implies dim(KerF)=n-1$ and this means that the imagine of the map $F$ is generated by a vector $u$ such that $ImF=\langle u \rangle$.
For this reason, $\forall v\in V, F(v)\in \langle u \rangle\implies F(v)=\lambda u$, for $\lambda \in \mathbb K$.
This also shows that the composition $(F\circ F)(v)=F(F(v))=F(\lambda u)=\lambda F(u)$, which is still an element of $\langle u \rangle$.
Since the Kernel of the function is the eigenspace associated to te eigenvalue $t=0$, it could be interesting studying the diagonal form of the representative matrix of the map.