I have a matrix:
$$A = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1\\ 0 & 1 & 1 & \cdots & 1\\ 0 & 0 & 1 & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}$$
I need to find $A^{-1}$. How could I do that, knowing that matrix A has size $n \times n$?
On
Think of the row operations that you would need to do to reduce this matrix to the identity matrix. Starting with $A$, the way that you would force the first row to be $\begin{bmatrix} 1 & 0 & \cdots & 0\end{bmatrix}$ would be to subtract the second row from the first row (in other words, replace the first row with the first row minus the second row). Similarly, to make the $i$th row have zeroes everywhere except in the $i$th entry (which is what the $i$th row of the identity matrix looks like), we should replace the $i$th row of $A$ with the $i$th row of $A$ minus the $(i + 1)$th row of $A$.
Thus, the matrix of row operations that reduces $A$ to the identity is the matrix $$ \begin{bmatrix} 1 & -1 & 0 & \cdots & \cdots & 0 \\ 0 & 1 & -1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & -1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \cdots & \cdots & 1 \end{bmatrix}$$
Note that we did not have to change the $n$th row, as it was already equal to the $n$th row of the identity matrix. Of course, the matrix that reduces $A$ to the identity matrix is $A^{-1}$, so indeed the matrix above is $A^{-1}$.
HINT: Consider the matrix
$$B = \begin{pmatrix} 1 & -1 & 0 &\cdots & 0\\ 0 & 1 & -1 & \cdots & 0\\ 0 & 0 & 1 & \ddots&0\\ \vdots & \vdots & \ddots & \ddots & -1\\ 0 & 0 & 0 & \cdots &1 \end{pmatrix}$$
One easily verifies that $$AB = I_n$$ and $$BA = I_n$$
Thus $B=A^{-1}$
Edit: How to come up with $B$? First of all you should compute some inverse matrices, maybe for the cases $n = 2$ and $n=3$. In order you may see a pattern and get an idea how a candidate $B$ for an inverse of $A$ should look like. Just verify $BA = I_n$ and $AB = I_n$ and you will end up with $A^{-1} = B$.