prove that an nxn nilpotent matrix always has det=0

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prove that an nxn nilpotent matrix always has det=0

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I know that at some point A^n=0, and therefore the determinant is zero, but I must also have to prove that the matrices before it becomes zero have a determinant of zero
and det(0)=0=A^n....and A^n is connected to A somehow
use the fact det[A^n] = [det A]^n

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ok, so that would conclude the proof, because only 0^n can give zero
just like x^n = 0, the only solution is x must be 0
k thanks!

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