Matrix Exponential for Nilpotent Matrix I want to prove that Id-A=${ e }^{ -{ \sum _{ k=1 }^{ n }{ \frac { { A }^{ k } }{ k }  }  } }$
whereas ${ A }^{ n }=0$ because A is nilpotent.
What i did was
setting z(t)=${ e }^{ -{ \sum _{ k=1 }^{ n }{ \frac { { At }^{ k } }{ k }  }  } }$
than obviously, we have  z(0)=Id and
z'(t)=$-\sum _{ k=1 }^{ n }{ { A }^{ k }{ t }^{ k-1 }{ e }^{ -{ \sum _{ k=1 }^{ n }{ \frac { { A }^{ k } }{ k }  }  } } } =-A\sum _{ k=0 }^{ n-1 }{ { (At) }^{ k }exp( } -\sum _{ k=1 }^{ n }{ \frac { { (At) }^{ k } }{ k }  }) $=$-A{ (Id-At) }^{ -1 }z(t)$.
Now we obtained a Differential equation.
z'(t)=$-A{ (Id-At) }^{ -1 }z(t)$
z(0)=Id
One can easily conclude now that  z(t)=(Id-At) is a solution, hence because of the uniqueness of solutions we get 
${ e }^{ -{ \sum _{ k=1 }^{ n }{ \frac { { A }^{ k } }{ k }  }  } }$=(Id-At)
which finishes the proof. Is this proof correct? And if it is are there any points which could be done better?
 A: You still have some typos, $A$ should be $A^k$ in the definition of $z(t)$, etc.

On the correctness, you only need to note that ${\rm Id}-At$ is invertible for small $t$ (which is sufficient for your purposes).

Also, depending on what you have learned (or on what you can refer to) you may need to consider instead the functions $z(t)c$ with $c\in\mathbb R^m$ when $A$ is $m\times m$. This would give equations on $\mathbb R^m$ and then it suffices to note that if two matrices $B,C$ satisfy $Bc=Cc$ for all $c$, then $B=C$.
A: Not an answer but a long comment that I think useful.
Identity :
$$I-A =\exp\left(-\sum_{k=1}^n \frac{1}{k}A^k\right) \tag{1}$$
seems to have fallen from the sky...
In fact, motivation for (1) is to be found on the somewhat equivalent identity :
$$\log(I-A)=-\sum_{k=1}^n  \frac{1}{k}A^k \tag{2}$$
where one recognizes the analog of the classical series expansion :
$$\log(I-z)=-\sum_{k=1}^{\infty}  \frac{1}{k}z^k$$
Of course, (2) is not equivalent to (1). In particular, there are strict conditions of convergence for the RHS (spectral radius < 1) limiting its application.
