Suppose $T^{k+1}=0$. Show that $I + T + T^2 + \dots+ T^k = (I - T)^{-1}.$ $T^{k+1} = 0$, where $k$ is a positive integer. Prove that $I + T + T^2 + \dots+ T^k = (I - T)^{-1}.$
My proof: Let $T$ be a $m \times n$ square matrix with the same number of $m$ rows and $n$ columns. $K$ must be greater than zero. 
$$I = I^2 * I^{-1}$$
$$T = T^2 * T^{-1}$$
$$T^{k-1} = T^k * T^{-1}$$
$$I-T^{-1} = \frac{1}{I-T}$$
$$I + T + T^2 + \dots + T^k = (I-T)^{-1}$$
$$I + T + T^k + T^2 + \dots + T^{k-1} = (I-T)^{-1}$$
$$I + T + T^k + T^2 + \dots + T^kT^{-1} = (I-T)^{-1}$$
I'm a bit stuck. It seems like one of those $P*P^{-1}$ problems.
Is what I am doing overkill to the proof?
I need directions. 
 A: There's a serious mistake in your approach: namely, that there is any such thing as $T^{-1}$. There isn't, since $T$ is nilpotent, and so has $0$ as an eigenvalue (its only one, actually). Also, I think you mean $m\times m$ or $n\times n$.
Ultimately, you're making things too complicated. Call $M=I+T+T^2+\cdots+T^k.$ Note that $$TM=T+T^2+T^3+\cdots+T^{k+1}=T+T^2+\cdots+T^k=M-I,$$ by assumption, and so we can readily see that $$(I-T)M=IM-TM=M-(M-I)=I.$$ Hence, $M=(I-T)^{-1},$ and we're done.
A: $$
\begin{align*}
&I + T + T^2 + \ldots + T^k  = (I - T)^{-1}\\
\iff & (I+ T + T^2 + \ldots + T^k )(I - T) = (I - T)^{-1}(I-T) = I\\
\iff & (I+ T + T^2 + \ldots + T^k ) - (T + T^2 + \ldots + T^k + T^{k+1}) = I\\
\iff & I+ (T + T^2 + \ldots + T^k ) - (T + T^2 + \ldots + T^k) - T^{k+1} = I\\
\iff & I - T^{k+1} = I\\
\iff & I - [0] = I
\end{align*}
$$
A: The identity for real numbers....
$$1+x+x^2+...+x^k=\frac{1-x^{k+1}}{1-x}$$
holds for matrices in the form
$$1+T+T^2+...+T^{k}=\left(1-T\right)^{-1}\,\left(1-T^{k+1}\right)$$. It is easy to show by multiplying both side by the same quantities
\begin{eqnarray}
(1-T)\,(1+T+T^2+...+T^{k})&=&1-T+(1-T)\,T+...+(1-T)\,T^{k}\\
&=&1-T+T-T^2+T^2-...-T^{k+1}=1-T^{k+1}
\end{eqnarray}
hence:
$$
(1+T+T^2+...+T^{k})=(1-T)^{-1}\,(1-T^{k+1})
$$
In your case $T^{k+1}=0$. 
A: Hint:
$$I + T + T^2 + ...+ T^k = \frac{1-T^{k+1}}{I-T}$$
