Prove that upper triangular matrices are closed under inversion Prove that the group of upper triangle matrices $G\subset GL_n(F)$ is also a subgroup of $GL_n(F)$
Contains identity: $I_n\in G$ since $I_{ij}=0$ when $i>j$
Closed under multi: $C=AB, c_{ij}=\sum_{k=1}^na_{ik}b_{kj},i=1,2..,n,j=1,2..,n$ for all $c_{ij},i>j$ either $i\ge k>j$ or $i>k\ge j$. One factor in each addend is $0$: either $a_{ik}=0$, or $b_{kj}=0$ thus $c_{ij}=0$ if $i>j\rightarrow C\in G$ 
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^{-1}\in G$ How do I know there is not a fringe case where $AA^{-1}=I, A^{-1}\not\in G$

Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0\subseteq V_1\subseteq V_2\subseteq\ldots\subseteq V_n$, where $\dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)\subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$.  The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
