Just to show a different approach.
Consider the matrix $\mathbf E$, having $1$ only on the first subdiagonal
$$
\mathbf{E} = \left\| {\,e_{\,n,\,m} = \left\{ {\begin{array}{*{20}c}
1 & {n = m + 1} \\
0 & {n \ne m + 1} \\
\end{array} } \right.\;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - 1 \\
\end{gathered} \right)\;} \right\|
$$
Multiply it by $\mathbf A$ , and it is easily seen that $$\mathbf E \, \mathbf A=\mathbf A-\mathbf I \quad \Rightarrow \quad \left( \mathbf I -\mathbf E \right)\,\mathbf A=\mathbf I$$
In another way, consider that the powers of $\mathbf E$ are readily found and have a simple formulation
$$
\begin{gathered}
\mathbf{E}^{\,\mathbf{2}} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered}
0 \\
n - k - 1 \\
\end{gathered} \right)\left( \begin{gathered}
0 \\
k - m - 1 \\
\end{gathered} \right)} \;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - 2 \\
\end{gathered} \right)\;} \right\| = \left\| {\,\left\{ {\begin{array}{*{20}c}
1 & {n = m + 2} \\
0 & {n \ne m + 2} \\
\end{array} } \right.\;} \right\| \hfill \\
\quad \vdots \hfill \\
\mathbf{E}^{\,\mathbf{q}} = \mathbf{E}^{\,\mathbf{q} - \mathbf{1}} \,\mathbf{E} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered}
0 \\
n - k - \left( {q - 1} \right) \\
\end{gathered} \right)\left( \begin{gathered}
0 \\
k - m - 1 \\
\end{gathered} \right)} \;} \right\| = \hfill \\
= \left\| {\,\left( \begin{gathered}
0 \\
n - m - q \\
\end{gathered} \right)\;} \right\|\quad \left| {\;0 \leqslant \text{integer}\;q} \right. \hfill \\
\end{gathered}
$$
Therefore
$$
\mathbf{A} = \sum\limits_{0\, \leqslant \,j} {\mathbf{E}^{\,\mathbf{j}} } = \frac{\mathbf{I}}
{{\mathbf{I} - \mathbf{E}}}
$$
To connect to the answer of Jean Marie, note that
$$
\left( {\mathbf{I} - \mathbf{E}} \right)\left\| {\,\begin{array}{*{20}c}
{x_{\,0} } \\
{x_{\,1} } \\
\vdots \\
{x_{\,n} } \\
\end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
{x_{\,0} ( - 0)} \\
{x_{\,1} - x_{\,0} } \\
\vdots \\
{x_{\,n} - x_{\,n - 1} } \\
\end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
{\nabla x_{\,0} \;\left| {x_{\, - 1} = 0} \right.} \\
{\nabla x_{\,1} } \\
\vdots \\
{\nabla x_{\,n} } \\
\end{array} \;} \right\|
$$