Given A is a nil-potent matrix (given $ A^k=0 $), prove that A-I is invertible. Is my proof correct? 
Given $A$ is a nil-potent matrix (given $A^k=0$), prove that A-I is invertible. 

I have proved the statement using contradiction, and I want to know if it is correct:
Let $ A-I \neq I.$  
Multiplying by $ A^{k-1} $ on both sides , we get:
$$A^k - A^{k-1} \neq A^{k-1},$$ but since $A^k=0$,
we get $ -A^{k-1} \neq A^{k-1}.$ Multiplying by A on both sides on left we get $0 \neq 0$, which is a contradiction. 
Is this a correct way of proving this? I have a doubt because we are basically multiplying 0 on both sides, without showing it explicitly here. 
 A: So consider following equation: $$I = I-\underbrace{A^k}_0 = (I+A+A^2+\ldots+A^{k-1})\cdot (I-A)$$ Therefore $(I-A)$ is invertible because we have an explicit inverse:
$$(I-A)^{-1} = I+A+A^2+\ldots+A^{k-1}$$
or
$$(A-I)^{-1} = -(I+A+A^2+\ldots+A^{k-1})$$
A: This does not work, because you don't have cancelation in the world of matrices: just because $AC=BC$ does not mean that $A=B$, even if $C\neq 0$.  However, here is a hint that might get you to a proof.
Consider the power series $$\frac{1}{1-x}=1+x+x^2+\ldots$$
A: The proof is incorrect. For one thing, $A - I \neq I$ is not the negation of $A- I $ being invertible. For another thing, it is unclear if multiplying by $A^{k-1}$ preserves the inequality. 
There are several ways to show what you are trying to show. One is to observe that $A$, being nilpotent, cannot have $1$ as eigenvalue, another would be to simply write down the inverse to this end start by factoring $I - A^k$. 
A: Let A is nilpotent matrix, s.t A^n=0 for some n
Consider another matrix 
$\sum_{j=0}^{n-1}-A^{j}$
multiply these two martices,
$(A-I)\sum_{j=0}^{n-1}-A^j$ =$\sum_{j=0}^{n-1}-A^{j+1}+\sum_{j=0}^{n-1}A^j$
$=-\sum_{j=1}^{n}A^{j}+\sum_{j=0}^{n-1}A^j$
$=-A^n-\sum_{j=1}^{n-1}A^{j}+\sum_{j=0}^{n-1}A^j+A^0$
$=I$
Therefore $\sum_{j=0}^{n-1}-A^{j}$ is inverse for A-I
