Inverse of an invertible triangular matrix (either upper or lower) is triangular of the same kind How can we prove that the inverse of an upper (lower) triangular matrix is upper (lower) triangular?

Let $$
A=\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1,n-1}     & a_{1,n}\\
      & a_{22} & \cdots & a_{2,n-1}     & a_{2,n}\\
      &       & \ddots & \vdots        & \vdots\\
      &       &        & a_{n-1,n-1} & a_{n-1,n}\\
      &       &        &               & a_{n,n}
\end{pmatrix}.
$$
Let $i,j$ be two integers such that $i,j\in\{1,\dots,n\}
$ and $i<j$.
Let $A_{i,j}$ be an $n-1\times n-1$ matrix which is obtained by crossing out row $i$ and column $j$ of $A$.
Then, $A_{i,j}$ is
$$
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1,i-1}     & a_{1,i} &a_{1,i+1}&a_{1,i+2}&\cdots&a_{1,j-1}&a_{1,j+1}&a_{1,j+2}&\cdots&a_{1n}\\
      & a_{22} & \cdots & a_{2,i-1}     & a_{2,i} &a_{2,i+1}&a_{2,i+2}&\cdots&a_{2,j-1}&a_{2,j+1}&a_{2,j+2}&\cdots&a_{2n}\\
      &       & \ddots & \vdots        & \vdots &\vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots\\
      &       &        & a_{i-1,i-1} & a_{i-1,i} &a_{i-1,i+1}&a_{i-1,i+2}&\cdots&a_{i-1,j-1}&a_{i-1,j+1}&a_{i-1,j+2}&\cdots&a_{i-1,n}\\
      &       &        &               & 0 & a_{i+1,i+1}&a_{i+1,i+2}&\cdots&a_{i+1,j-1}&a_{i+1,j+1}&a_{i+1,j+2}&\cdots&a_{i+1,n}\\
      &       &        &               & & 0&a_{i+2,i+2}&\cdots&a_{i+2,j-1}&a_{i+2,j+1}&a_{i+2,j+2}&\cdots&a_{i+2,n}\\
      &       &        &               & & &0&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots\\
      &       &        &               & & &&\ddots&a_{j-1,j-1}&a_{j-1,j+1}&a_{j-1,j+2}&\cdots&a_{j-1,n}\\
      &       &        &               & & &&&0&a_{j,j+1}&a_{j,j+2}&\cdots&a_{j,n}\\
      &       &        &               & & &&&&a_{j+1,j+1}&a_{j+1,j+2}&\cdots&a_{j+1,n}\\
      &       &        &               & & &&&&&a_{j+2,j+2}&\cdots&a_{j+2,n}\\
      &       &        &               & & &&&&&&\ddots&\vdots\\
      &       &        &               & & &&&&&&&a_{n,n}\\
\end{pmatrix}.
$$
So, $\det A_{i,j}=0$ if $i,j$ are two integers such that $i,j\in\{1,\dots,n\}
$ and $i<j$.
Let $C_{i,j}$ be the $(i,j)$-cofactor of $A$.
Then, $C_{i,j}=(-1)^{i+j}\det A_{i,j}=0$ if $i,j$ are two integers such that $i,j\in\{1,\dots,n\}
$ and $i<j$.
So, $$A^{-1}=\frac{1}{\det A}\begin{pmatrix}C_{11}&C_{21}&\cdots&C_{n,1}\\
C_{12}&C_{22}&\cdots&C_{n,2}\\
\vdots&\vdots&&\vdots\\
C_{1n}&C_{2n}&\cdots&C_{n,n}\\
\end{pmatrix}=\frac{1}{\det A}\begin{pmatrix}C_{11}&C_{21}&\cdots&C_{n,1}\\
0&C_{22}&\cdots&C_{n,2}\\
\vdots&\vdots&&\vdots\\
0&0&\cdots&C_{n,n}\\
\end{pmatrix}.$$
