<s>[INST] Find the inverse of a lower triangular matrix of ones Find the inverse of the matrix $A=(a_{ij})\in M_n$ where
$$
\begin{cases}
a_{ij}=1, &i\geq j,\\
a_{ij}=0, &i<j.
\end{cases}
$$
The only method for finding inverses that I was taught was by finding the adjugate matrix. So $A^{-1}=\frac{1}{\det A}\operatorname{adj(A)}$
$$A=\begin{pmatrix}
1 & 0 &0 &\ldots &0\\
1 & 1 & 0 &\ldots &0\\
1 & 1 & 1 &\ldots &0\\
\vdots &\vdots & \vdots &\ddots & \vdots\\
1 & 1 &1 & \ldots &1 \end{pmatrix}$$
This is a triangular matrix so $\det A=1.$
To find the adjugate I first need to find the cofactor matirx$(C)$ of $A$.
$$C=\begin{pmatrix}
1&-1&0&0&\ldots &0\\
0 & 1& -1 &0&\ldots & 0\\
0 & 0 & 1 & -1&\ldots &0\\
0 & 0 & 0 &1 &\ldots &0\\
\vdots &\vdots &\vdots &\vdots&\ddots & \vdots\\
0 & 0 &0 &0 &\ldots &1\end{pmatrix}$$
$$C^T=\operatorname{adj}(A)=\begin{pmatrix}
1 & 0 & 0 & 0&\ldots &0\\
-1 & 1& 0 & 0 &\ldots &0\\
0&-1&1&0&\ldots &0\\
0& 0 &-1 &1 &\ldots&0\\
\vdots &\vdots &\vdots &\vdots &\ddots&\vdots\\ 
0&0&0&0&\ldots &1\end{pmatrix}=A^{-1}$$
Is this correct? Also, can I leave it like that or should I somehow write it more formally?
 [/INST] 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 
$$
\left( {\mathbf{I} - \mathbf{E}} \right)\left\| {\,\begin{array}{*{20}c}
   {x_{\,0} }  \\
   {x_{\,1} }  \\
    \vdots   \\
   {x_{\,n} }  \\
 \end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
   {x_{\,0} ( - 0)}  \\
   {x_{\,1}  - x_{\,0} }  \\
    \vdots   \\
   {x_{\,n}  - x_{\,n - 1} }  \\
 \end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
   {\nabla x_{\,0} \;\left| {x_{\, - 1}  = 0} \right.}  \\
   {\nabla x_{\,1} }  \\
    \vdots   \\
   {\nabla x_{\,n} }  \\
 \end{array} \;} \right\|
$$
</s>