The claim is false without further conditions on the field. For example
$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\mathcal M_2(\overline{\Bbb F}_2)$
has $\,x^2-1=(x-1)^2\,$ as characteristic and minimal polynomial and thus $\,A^2=I\,$ , yet it even isn't diagonalizable...
Added as a result of an added condition: if $\,\operatorname {char}\Bbb F=0\,$ then all the roots of $\,x^n-1\,$ are simple (why?), and since the definition field $\,\Bbb F\,$ is algebraically closed then we can write
$x^n-1=\prod_{i}(x-a_i)$
where $\,a_i^n=1\,$ are $\,n\,$-th roots of unit in the field. Since clearly the characteristic and the minimal polynomial of $\,A\,$ coincide in this case (why?), the last one can be written as a product of different linear factors and thus $\,A\,$ is diagonalizable, with its eigenvalues, i.e. the roots of unit $\,a_i'\,$s, on its main diagonal.