Power sums determine the characteristic polynomial

Continuity. The entries of $A^k$ are polynomials in the entries of $A$, and the trace is linear. So each $\operatorname{tr}A^k$ is a polynomial function of $A$, hence continuous.

Link with the spectrum. Over $\mathbb{C}$, triangularise $A=PTP^{-1}$ with diagonal $\lambda_1,\dots,\lambda_n$ (with algebraic multiplicities). Then $A^k=PT^kP^{-1}$ and $T^k$ is triangular with diagonal $\lambda_i^k$, so
$$\operatorname{tr}A^k=\sum_i\lambda_i^k=:p_k.$$

Newton's identities. Write $\chi_A(t)=\prod_i(t-\lambda_i)=t^n-e_1t^{n-1}+e_2t^{n-2}-\dots+(-1)^ne_n$ where $e_k$ are the elementary symmetric polynomials of the $\lambda_i$. Newton's identities state that for $1\le k\le n$
$$ke_k=\sum_{j=1}^k(-1)^{j-1}e_{k-j}\,p_j,\qquad e_0=1.$$
Solving successively gives $e_1=p_1$, $e_2=\frac{p_1^2-p_2}2$, $e_3=\frac{p_1^3-3p_1p_2+2p_3}6$, and so on: each $e_k$ is a polynomial in $p_1,\dots,p_k$ with rational coefficients (dividing by $k$ is allowed in characteristic $0$).

Conclusion. The coefficients of $\chi_A$ are polynomials in $\operatorname{tr}A,\dots,\operatorname{tr}A^n$. Hence two matrices with the same first $n$ power traces have the same characteristic polynomial.

Check. $n=2$: $\chi_A(t)=t^2-(\operatorname{tr}A)t+\frac{(\operatorname{tr}A)^2-\operatorname{tr}A^2}2$, and indeed $\det A=\frac{(\operatorname{tr}A)^2-\operatorname{tr}(A^2)}2$.

Remark. The division by $k$ is essential: over a field of characteristic $p\le n$ the power sums do not determine the characteristic polynomial.