In the Jordan normal form of A, the diagonal entries will all be roots of unity, and the off-diagonal elements must all be zero (because when you take powers of it, the off-diagonal elements can only get bigger). So A has a diagonal JNF and is therefore diagonalisable.

Suppose that . Multiply both sides by and you see that , hence .

Therefore . Multiply both sides by and you see that . And so on. So all the coefficients are zero and thus the vectors are lin. ind.