$I+B$ is invertible if $B^{k} = 0$ If $B$ is nilpotent and $B^{k} = 0$ (and B is square), how should I go around proving that $I + B $ is invertible? I tried searching for a formula - $I = (I + B^{k}) = (I + B)(???)$
But I didn't get anywhere :(

Essentially if $B$ is idempotent of order $n$ you can find a basis where 
$$B=\left(\begin{array}{ccccc}
0 & b_{12} & 0 & \cdots & 0\\
0 & 0 & b_{23} &  & 0\\
0 & 0 & 0 &  & 0\\
\vdots &  &  & \ddots & b_{n-1n}\\
0 & 0 & 0 & \cdots & 0
\end{array}\right)$$
In this form you can easly see what's happening when you calculate $B^k$. For example squaring $B$ we get:
$$B^2=\left(\begin{array}{ccccc}
0 &0 & b_{12}b_{23} & 0 & \cdots & 0\\
0 &0 & 0 & b_{23}b_{34} &  & 0\\
0 &0 & 0 & 0 & \ddots &  \vdots \\
0 &0 & 0 & 0 & \ddots &  b_{n-2n-1}b_{n-1n}\\
\vdots&\vdots &  &  & \ddots &0 \\
0 &0 & 0 & 0 & \cdots & 0
\end{array}\right)$$
Everytime you multiply for $B$ you shift the non zero columns and rows by one. You can fill out the details yourself, (for example treating the general case of $B$ idempotent of order k, doesn't change much) this was just to give you an useful picture of what's happening....
Clearly in this base
$$I+B=\left(\begin{array}{ccccc}
1 & b_{12} & 0 & \cdots & 0\\
0 & 1 & b_{23} &  & 0\\
0 & 0 & 1 &  & 0\\
\vdots &  &  & \ddots & b_{n-1n}\\
0 & 0 & 0 & \cdots & 1
\end{array}\right)$$
and $det(I+B)=1$ and so it's clearly invertible. 
In the general case you can always write $I+B$ as un upper triangular matrix with $1$ on the diagonal and so with $det(I+B)=1$
