Traces of all positive powers of a matrix are zero implies it is nilpotent Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be $0$, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.
What I would like to show is the converse, that is, 

if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.

I tried to show that $0$ must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.
May I know of the approach to show that $A$ is nilpotent?

If the eigenvalues of $A$ are $\lambda_1$, $\dots$, $\lambda_n$, then the eigenvalues of $A^k$ are $\lambda_1^k$, $\dots$, $\lambda_n^k$. It follows that if all powers of $A$ have zero trace, then $$\lambda_1^k+\dots+\lambda_n^k=0\qquad\text{for all $k\geq1$.}$$ Using Newton's identities to express the elementary symmetric functions of the $\lambda_i$'s in terms of their power sums, we see that all the coefficients of the characteristic polynomial of $A$ (except that of greatest degree, of course) are zero. This means that $A$ is nilpotent.
