 Solution

Let column $j$ of $A^{-1}$ be $x$, so $Ax=e_j$. Every diagonal entry of the invertible lower triangular matrix $A$ is nonzero, since its determinant is the product of those entries. In row one through row $j-1$, the right-hand side is zero. Forward substitution first gives $x_1=0$, then $x_2=0$, and successively $x_i=0$ for all $i<j$. These are precisely the entries above the diagonal in column $j$ of $A^{-1}$. Every column satisfies this, proving that $A^{-1}$ is lower triangular. This is triangularity of an inverse matrix.