If $ \operatorname{Tr}(M^k) = \operatorname{Tr}(N^k)\;\forall 1\leq k \leq n$, then how do we show the $M$ and $N$ have the same eigenvalues? Let $M,N$ be $n \times n$ square matrices over an algebraically closed field with the properties that the trace of both matrices coincides along with all powers of the matrix. More specifically, suppose that $\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)$ for all $1\leq k \leq n$. The following questions about eigenvalues is then natural and I was thinking it would be an application of Cayley-Hamilton but I am having trouble writing out a proof.
How do we show that $M$ and $N$ have the same eigenvalues?
Added (because this question is now target of many duplicates, it should state its hypotheses properly). Assume that all the mentioned values of $k$ are nonzero in the field considered; in other words either the field is of characteristic $0$, or else its prime characteristic $p$ satisfies $p>n$.
 A: What about this one?
We know that if $\psi_A(x) = |xI_n - A| = x^n+c_{n-1}x^{n-1}+c_{n-2}x^{n-2}+\cdots+c_1x+c_0I_n$ be characteristic polynomial of $A$ (thus $c_0 = (-1)^n|A|$), then the coefficients are given by 
$$c_{n-m}=\frac{(-1)^m}{m}\left|
                              \begin{array}{ccccc}
                                t_1 & m-1 & 0 & \cdots &  0 \\
                                t_2 & t_1 & m-2 & \cdots &  0  \\
                                \vdots  & \vdots &   &   &  \vdots \\
                                t_{m-1} & t_{m-2} & t_2 & t_1 & 1 \\
                                t_m & t_{m-1} & t_3 & t_2 & t_1 \\
                              \end{array}
                            \right|$$ 
where $t_r:= \operatorname{tr}(A^r)$.
Assume that $\psi_{M}(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1x+a_0I_n$ and $\psi_N(x)=x^n+b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_1x+b_0I_n$. Then $$a_r=b_r$$ (by using $\operatorname{tr}(M^k) = \operatorname{tr}(N^k)$ and the above determinant). 
Hence $\psi_M(x)=\psi_N(x)$ which means $M$ and $N$ will have same eigenvalues.
A: $M$ and $N$ have the same generalized eigenvalues (with multiplicity) if and only if their characteristic polynomials are the same. Thus it suffices to show that
the power sums $T_k = \sum _{i=1}^{n} \alpha^k_i$ for $k = 1$ to $n$ generate the ring of symmetric polynomials (the coefficients of the characteristic polynomial
are symmetric polynomials in the eigenvalues). This is a result due to Newton.
For example, if $S_k$ is the usual $k$-th elementary symmetric polynomial
(the sum of all products of $k$ distinct $\alpha_i$), then one has:
$$S_1 = \sum \alpha_i =  T_1,$$
$$S_2 = \sum_{i > j} \alpha_i \alpha_j = \frac{1}{2}
\left( \left(\sum \alpha_i\right)^2 - \sum \alpha^2_i  \right)
 =
\frac{1}{2}(T^2_1 - T_2)$$
More generally, one has:
$$\log \sum_{k=0}^{n} S_k X^k = \log \prod_{i=1}^{n} (1 + \alpha_i X)
= \sum_{k=1}^{n} \log(1 + \alpha_i x)$$
which, expanding the logarithm, becomes:
$$
\sum_{k=1}^{n} \sum_{j=1}^{\infty} \frac{\alpha^j_i (-1)^{j-1} X^j}{j}
= \sum_{j=1}^{\infty} \frac{T_j (-1)^{j-1} X^j}{j}$$
In particular, from the $S_k$ one can determine all the $T_k$, and from
the $T_k$ (for $k = 1$ to $n$) one can determine the $S_k$ (and hence all the $T_k$ as well). 
This even shows that the ring generated by $T_k$ over any ring $R$ for $k = 1$ to $n$ is the same
as the ring generated by $S_k$ for $k = 1$ to $n$, as long as $n!$ is invertible in $R$. So the result also holds for any field of characteristic $p > n$.
It is false if $p \le n$ - for example, the identity $p \times p$
matrix and all its powers has trace $0$, which is the same as the zero matrix.
