TITLE: Proving that the coefficients of the characteristic polynomial are the traces of the exterior powers
QUESTION [23 upvotes]: 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!

REPLY [4 votes]: 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,