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?

Here is an argument that does not involve Newton's identities, although it is still closely related to symmetric functions. Write
$$f(z) = \sum_{k\ge 0} z^k \text{tr}(A^k) = \sum_{i=1}^n \frac{1}{1 - z \lambda_i}$$
where $\lambda_i$ are the eigenvalues of $A$. As a meromorphic function, $f(z)$ has poles at the reciprocals of all of the nonzero eigenvalues of $A$. Hence if $f(z) = n$ identically, then there are no such nonzero eigenvalues. 
The argument using Newton's identities, however, proves the stronger statement that we only need to require $\text{tr}(A^k) = 0$ for $1 \le k \le n$. Newton's identities are in fact equivalent to the identity
$$f(z) = n - \frac{z p'(z)}{p(z)}$$
where $p(z) = \prod_{i=1}^n (1 - z \lambda_i)$. To prove this identity it suffices to observe that
$$\log p(z) = \sum_{i=1}^n \log (1 - z \lambda_i)$$
and differentiating both sides gives
$$\frac{p'(z)}{p(z)} = \sum_{i=1}^n \frac{- \lambda_i}{1 - z \lambda_i}.$$
(The argument using Newton's identities is also valid over any field of characteristic zero.) 
