Let $\Phi$ an endomorphism of a finite dimensional vector space on an algebraically closed field $K$. Suppose that $\Phi$ has finite order and that $Char K$ not divide the order of $\Phi$. Prove that $\Phi$ is diagonalizable.
My idea is:
Let $A$ a Jordan matrix associated to the endomorphism $\Phi$, so exists $D$ diagonal and $N$ nilpotent such that $$A=D+N.$$ if we call $\alpha$ the order of $\Phi$, we obtain $$A^\alpha=Id.$$ Clearly the eigenvalues of $A$ are $\alpha$-th roots of unity and the same holds for $D$. So $$Id=A^\alpha=(D+N)^\alpha=\sum_{i=0}^\alpha \binom {\alpha}{i} D^{\alpha-i}N^i=D^{\alpha}+\sum_{i=i}^n \binom {\alpha}{i} D^{\alpha-i}N^i =Id +\sum_{i=i}^n \binom {\alpha}{i} D^{\alpha-i}N^i $$ then $$\sum_{i=i}^n \binom {\alpha}{i} D^{\alpha-i}N^i=0$$ and since $CharK$ not divide $\alpha$ we have that $\binom {\alpha}{i}$ is not always $0$.
Now I want to conclude that $N=0$ but I don’t know hot to use the property of the matrix $N$.
I must say I prefer Matthew's approach. That said, your own approach does work.
To follow through with your approach, it might be helpful to build intuition by writing down what $D$ and $N$ look like in an explicit example.
Let's work through the example where $A$ has a single Jordan block, of dimension three. Some simple calculation gives \begin{align} A & = D + N \\ & = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} + \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \\ \\ A^2 & = D^2 + 2DN + N^2 \\ & = \begin{bmatrix} \lambda^2 & 0 & 0 \\ 0 & \lambda^2 & 0 \\ 0 & 0 & \lambda^2 \end{bmatrix} + 2 \begin{bmatrix} 0 & \lambda & 0 \\ 0 & 0 & \lambda \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \\ A^3 & = D^3 + 3D^2N + 3DN^2 + N^3 \\ & = \begin{bmatrix} \lambda^3 & 0 & 0 \\ 0 & \lambda^3 & 0 \\ 0 & 0 & \lambda^3 \end{bmatrix} + 3 \begin{bmatrix} 0 & \lambda^2 & 0 \\ 0 & 0 & \lambda^2 \\ 0 & 0 & 0 \end{bmatrix} + 3 \begin{bmatrix} 0 & 0 & \lambda \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{align} ... and so on.
Now ask yourself: In this example I just gave you, is it possible for $A^\alpha$ to be equal to $I$, for any $\alpha \geq 1$ such that $\text{char}(K) \nmid \alpha$?
And then, having satisfied yourself with this example, perhaps you could convince yourself that the only way that you can get $A^\alpha = I$ is if all the Jordan blocks of $A$ have dimension one.
I'd also remark that the statement that "a matrix is diagonalisable iff its minimal polynomial has simple roots" is trivial to prove once you know that Jordan decompositions exist. (All of this assumes $K$ is algebraically closed.)
Again, I'll help you build intuition through examples. Consider the following diagonal matrix. $$ A_1 = \begin{bmatrix} \lambda_1 & 0 & 0 & 1 \\ 0 & \lambda_1 & 0 & 0 \\ 0 & 0 & \lambda_2 & 0 \\ 0 & 0 & 0 & \lambda_3 \end{bmatrix}.$$
Its minimal polynomial is $$ m_{A_1}(X) = (X - \lambda_1) (X - \lambda_2)(X - \lambda_3),$$ which has only simple roots.
On the other hand, consider the following Jordan matrix, which is non-diagonal. $$ A_2 = \begin{bmatrix} \lambda_1 & 1 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 \\ 0 & 0 & \lambda_1 & 0 \\ 0 & 0 & 0 & \lambda_2 \end{bmatrix}.$$
This matrix has minimal polynomial $$ m_{A_2}(X) = (X - \lambda_1)^2 (X - \lambda_2),$$ which has a double root at $\lambda_1$.
Perhaps think about how this pattern generalises.