Let $A, B \in M_n(\mathbb{H})$ such that $AB=I$. Why does it follow that $BA=I$?
All the proofs I found of this fact for matrices over a field relied either on the determinant of a matrix (which is not defined for a quaternionic matrix) or on dimension of the vector space the matrices act upon (which is not applicable since $\mathbb{H}^n$ is not a vector space).
I asked a less focused question a few days ago (where you can also see the context for why I am wondering this) and I was given some hints that rely on algebraic facts which I am not familiar with. This is the last piece in the puzzle to finish a proof in a book which has no algebraic prerequisites beyond basic linear algebra, so I would much prefer to get an elementary answer, but if no such answer is possible, I would love to also get explanations for the algebraic terms used and why does this follow from them.
The determinant is defined over any field, and $\mathbb H^n$ is a vector space (over $\mathbb Q$, over $\mathbb R$, over $\mathbb C$). So any proof that uses those two facts will work the same in $M_n(\mathbb H)$.
For another view, quaternions can be represented as (certain) $4\times 4$ real matrices, so $M_n(\mathbb H)$ can be seen as $M_{4n}(\mathbb R)$ and you can use the proof you are comfortable with.