I am trying assignment questions of Linear algebra and couldn't solve this particular question regarding diagonalizabilty.

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> Let $n \times n$ complex matrix $A$ satisfies $A^k = I$ the $n \times n $ identity matrix, where $k$ is a positive integer $>1$ and let $1$ not be an eigenvalue of $A$. Then how to prove that A is necessarity Diagonalizable?
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As $A^k=I$ and 1 is not an eigenvalue so $(A-I ) (A^{k-1}+...+ I)=0$ implies that $(A^{k-1}+...+ I)=0$ but I am unable to move foreward .

Can you please help?