In a finite dimensional vector space let $A,B$ be square matrices then $AB = I$ implies $BA = I$.
My idea for a proof goes like this: $BAx = BABy = By = x$. So I need to show that forall $x$ exists $y$ with $x = By$.
The reason this is true is that since $A,B : k^N \to k^N$ are linear maps between finite spaces they have a similar property to function on finite sets: An injection is a bijection. A surjection is a bijection.
To prove this we can show that, if $B$ is injective, it takes a basis to another basis:
Let $\{e_i\}$ be a basis then set $f_i = B e_i$. Suppose we have a linear relation $\sum c_i f_i = 0$ then we will show that each $c_i = 0$.
What we have is $\sum c_i B e_i = 0$ or $B \sum c_i e_i = 0$ and by injectivity we this gives $\sum c_i e_i = 0$ which gives each $c_i = 0$ as $\{e_i\}$ is a basis.
Now $AB = I$ so $\ker(AB) = 0$ implies $\ker(B) = 0$, by the above we know $B$ is a bijection so for all $x$ there exists $y$ such that $x = B y$.