Nilpotent Matrices: Show Jordan Form w/Linear Independence

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Homework Statement


Suppose that N is a nilpotent mxm matrix, N^{m}=0, but N^{m&#039;}\neq0 for m'<m. Show that there exists a basis in which it takes the form of a single Jordan block with vanishing diagonal elements. Prove that your basis set is linearly independent.


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The Attempt at a Solution


So I've recognized the fact that N^{m&#039;}\neq0 for m'<m means that N^{m&#039;} does not annihilate every vector in V. I'm just not really sure where go from here...
 
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Pick a basis for the vectors that are annihilated by N. Then add a basis for the vectors that are annihilated by N^2 but not by N. Continue. Eventually you'll get a basis for the whole space, right? What does N look like in that basis?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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