Proving that the coefficients of the characteristic polynomial are the traces of the exterior powers Let $T$ be an endomorphism of a finite-dimensional vector space $V$. Let $$f(x)=x^n+c_1x^{n-1}+ \dots + c_n$$ be the characteristic polynomial of $T$. It is well known that $c_m=(-1)^m\text{tr}(\bigwedge^mT)$. 
If the base field is $\mathbf{C}$, then we can prove it using a density argument. The statement is true for diagonalizable matrices, which are dense in $M_n(\mathbf{C})$. This actually enough to prove it general, but I don't find it very illuminating. I would like to see an abstract proof of this result.
Thank you!
 A: I know this question is quite old, but still I was asking myself the same question recently so I was trying to find the solution ; finding this question (and Feanor's answer) wasn't completely enlightening, although it was helpful. I took the time to type my solution up, hence I thought I should add it here. 
There is a little intro to exterior algebra because I admit I wasn't totally familiar with it myself and I had to read back through my notes. I assume not every reader will be familiar with it so I thought it was a good idea. It's on my academia.edu page, which I will link here. 
https://www.academia.edu/5901122/On_The_Characteristic_Polynomial
Note : the question suggests it holds only for endomorphisms of finite dimensional vector spaces, but in fact it works for any endomorphism of $R^n$ where $R$ is a commutative unital ring. There's really nothing special about $\mathbb C$ in the statement.
Hope that helps,
A: $\newcommand{\tr}{\operatorname{tr}}$ 
First, let us give an alternative description of $\tr {\bigwedge}^m T$.
Let $e_1,e_2,\dots,e_n$ be a fixed basis for $V$, and for $J \subset \{1,2,\dots,n\}$ with $|J| = m$, let $T_J$ denote the $m \times m$ matrix acquired by keeping the entries $T_{i,j}$ with $i,j \in J$. Alternatively, if $P_J$ is the projection onto $\operatorname{span} \{e_i\}_{i \in J}$ then $T_J = P_J T P_J$. My claim is that:
$$\tr {\bigwedge}^m T = \sum_J \det T_J$$ where the sum runs over all $J$ with $|J| = m$. More precisely, if $e_J := \bigwedge_{i \in J} e_i$ (in any order, but fixed once and for all), then ${\bigwedge}^m T e_J = \det T_J e_J + \sum_{I \neq J} \alpha_I e_I$. This can be showed by brute-force application of the definition on exterior power and the determinant (the one with sum over permutations). Another possibility is to say that:
$$ 
{\bigwedge}^m P_J {\bigwedge}^m T e_J
 = {\bigwedge}^m P_J {\bigwedge}^m T {\bigwedge}^m P_J e_J 
 =  {\bigwedge}^m T_J e_J = \det T_J e_J
$$
where the last equality holds because $m$ is the dimension of the space on which $T_J$ acts (Wiki lists this property as one of the possible definitions of $\det$).
Now, because $\tr {\bigwedge}^m T$ is by definition the sum of coeffs at $e_J$ in ${\bigwedge}^m T e_J$ (note that $e_J$ form the basis), it turns out that the formula with the sum works.
Thus, it remains to see that $c_m = (-1)^m \sum_J \det T_J$. This can be done by looking at how the characteristic polynomial is computed from the definition, using the matrix form of $T$.  I think this is known, and has been asked on MSE (except the sign differs). The gist of the proof I know is that to get $(-x)^{n-m}$ from $\det(T - xI)$, you need to choose the term $-x$ in $n-m$ places, and from what remains you get the coefficient $\det T_J$, where $J$ is the set of indices where you did not take the $-x$.
