Let's write $L^{-1}=[y_1\:\cdots\:y_n],$ where each $y_k$ is an $n\times 1$ matrix.
Now, by definition, $LL^{-1}=I=[e_1\:\cdots\:e_n],$ where $e_k$ is the $n\times 1$ matrix with a $1$ in the $k$th row and $0$s everywhere else. Observe, though, that $LL^{-1}=L[y_1\:\cdots\:y_n]=[Ly_1\:\cdots\: Ly_n],$ so $Ly_k=e_k\qquad(1\leq k\leq n)$
By the proposition, since $e_k$ has only $0$s above the $k$th row and $L$ is lower triangular and $Ly_k=e_k$, then $y_k$ has only $0$s above the $k$th row. This is true for all $1\leq k\leq n$, so since $L^{-1}=[y_1\:\cdots\:y_n],$ then $L^{-1}$ is lower triangular, too.
$********$
Here's an alternative (but related) approach.
Observe that a lower triangular matrix is nonsingular if and only if it has all nonzero entries on the diagonal. Let's proceed by induction on $n$. The base case ($n=1$) is simple, as all scalars are trivially "lower triangular". Now, let's suppose that all nonsingular $n\times n$ lower triangular matrices have lower triangular inverses, and let $A$ be any nonsingular $(n+1)\times(n+1)$ lower triangular matrix. In block form, then, we have $A=\left[\begin{array}{c|c}L & 0_n\\\hline x^T & \alpha\end{array}\right],$ where $L$ is a nonsingular $n\times n$ lower triangular matrix, $0_n$ is the $n\times 1$ matrix of $0$s, $x$ is some $n\times 1$ matrix, and $\alpha$ is some nonzero scalar. (Can you see why this is true?) Now, in compatible block form, we have $A^{-1}=\left[\begin{array}{c|c}M & b\\\hline y^T & \beta\end{array}\right],$ where $M$ is an $n\times n$ matrix, $b,y$ are $n\times 1$ matrices, and $\beta$ some scalar. Letting $I_n$ and $I_{n+1}$ denote the $n\times n$ and $(n+1)\times(n+1)$ identity matrices, respectively, we have $I_{n+1}=\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right].$ Hence, $\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=I_{n+1}=A^{-1}A=\left[\begin{array}{c|c}ML+by^T & M0_n+b\alpha\\\hline x^TM+\alpha y^T & y^T0_n+\beta\alpha\end{array}\right]=\left[\begin{array}{c|c}ML+by^T & \alpha b\\\hline x^TM+\alpha y^T & \beta\alpha\end{array}\right].$ Since $\alpha$ is a nonzero scalar and $\alpha b=0_n$, then we must have $b=0_n$. Thus, $A^{-1}=\left[\begin{array}{c|c}M & 0_n\\\hline y^T & \beta\end{array}\right],$ and $\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=\left[\begin{array}{c|c}ML & 0_n\\\hline x^TM+\alpha y^T & \beta\alpha\end{array}\right].$ Since $ML=I_n$, then $M=L^{-1}$, and by inductive hypothesis, we have that $M$ is then lower triangular. Therefore, $A^{-1}=\left[\begin{array}{c|c}M & 0_n\\\hline y^T & \beta\end{array}\right]$ is lower triangular, too, as desired.