Prove that $A^5 \neq I$ $A \in M_{5}(\mathbb{C})$, $\operatorname{trace}(A) = 0$, $I-A$ is invertible. Prove that $A^5 \neq I$
I think this problem has something to do with eigenvalue and the fact that trace of a matrix = sum of the eigenvalues. But, I have no idea how to proceed!
Thanks! 

Suppose you had $A^5 = I$. Then the eigenvalues of $A$ are all fifth roots of unity. The fact that $I - A$ is invertible means that $1$ is not an eigenvalue. So you're left with the other $4$ fifth roots of unity. 
You're given that the trace of $A$ is zero, so the sum of all eigenvalues of zero. So if you can verify that there's no way of adding the remaining fifth roots of unity to get zero then you're done. But any such way would result in an equation
$$c_3e^{8\pi i \over 5} + c_2e^{6\pi i \over 5} +c_1e^{4\pi i \over 5} + c_0e^{2\pi i \over 5} = 0$$
Here the $c_i$ are integers. So $z = e^{2\pi i \over 5}$ satisfies 
$$c_3z^3 + c_2z^2 + c_1z + c_0 = 0 \tag{*}$$
But $x^4 + x^3 + x^2 + x + 1$ is the minimal polynomial of $z$ ( since cyclotomic polynomials are irreducible.) Hence $(*)$ cannot happen and thus the assumption that $A^5 = I$ leads to a contradiction.
