Recognizing Patterns in Alternating Signs Matricies and their Inverses Let's say we have the matrix A with alternating-sign 1's below
A = \begin{bmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{bmatrix}
If we find the inverse, we get 
A^-1 = \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{bmatrix}
We get a similar pattern for 5 x 5, 6 x 6, ..., n x n matricies. 
How would we prove that we would achieve this pattern for all inverses of n x n matricies? 

To elaborate on the other answer, which is correct, though presented with no concrete examples, note that your matrix $A$ is a sum $A = I + N$, where $N$ is the nilpotent matrix
$$N = \begin{bmatrix} 0 & 1 & & \\ & 0& 1 & \\ & & 0& 1 \\ & & &0 \end{bmatrix}$$
The matrix $N$ has powers of a simple form, for example
$$N^2 = \begin{bmatrix} 0 & 0 &1 & \\ & 0& 0 &1 \\ & & 0& 0 \\ & & &0 \end{bmatrix}, \quad N^3 = \begin{bmatrix} 0 & 0 &0 & 1\\ & 0& 0 &0 \\ & & 0& 0 \\ & & &0 \end{bmatrix}, \quad N^4 = 0$$
It is a basic fact in commutative algebra that when you have a nilpotent element $N$, then $1 + N$ is invertible, since (I'll suppose here that $N^4 = 0$) we have
$$I = I^4 + N^4 = (I + N)(I - N + N^2 - N^3)$$
and so $(I - N + N^2 - N^3)$ must be the inverse of $(I + N)$.
