$\operatorname{tr}(A)=\operatorname{tr}(A^{2})= \ldots = \operatorname{tr}(A^{n})=0$ implies $A$ is nilpotent

Two facts: (1) The trace of a matrix $A$ is the sum of it's eigenvalues and (2) the eigenvalues of $A^k$ are $\lambda^k$ where $\lambda$ is an eigenvalue of $A$ matched with algebraic multiplicities.

This means we can write $$\lambda_1 + \cdots + \lambda_n = 0$$ $$\lambda_1^2 + \cdots + \lambda_n^2 = 0$$ $$\cdots$$ $$\lambda_1^n + \cdots + \lambda_n^n = 0$$

It is also a known fact that all symmetric polynomials can be expressed as a polynomial in the above symmetric polynomial forms. The characteristic equation of $A$,

$$\det(A-\lambda I) = (\lambda - \lambda_1) \cdots (\lambda - \lambda_n) =\\ \lambda^n + f_{n-1}(\lambda_1, \dots, \lambda_n) \lambda^{n-1} + \cdots + f_0(\lambda_1, \cdots, \lambda_n),$$

has coefficients symmetric in $\lambda_1, \cdots, \lambda_n$. This means they must be constant as the expression in terms of the powers of eigenvalues can be evaluated with those expressions equal to zero. I.e., the coefficients are in fact not dependent on the eigenvalues at all! But since each of the coefficients will go to zero as $|\lambda_1 \cdots \lambda_n| \to 0$ the constant terms must be zero and hence the characteristic equation is $\lambda^n$ identically.

Use the Cayley-Hamilton theorem in combination with Newton's identities.

I think you can show that the characteristic polynomial $\det(\lambda I - A)$ of $A$ is in fact $\lambda^n$ using newtons formula on symmetric polynomials.