How to prove that a matrix is invertible $\iff$ invertible at right $\iff$ invertible at left? Let $A$ a $n\times n$ invertible at left. In fact, I just want to prove that it's invertible at right (the rest is obvious). All what I can say is that there is a $B$ s.t. $BA=I.$
To prova $AB=I$, I have problem. I have that $$AB=AB^2A=BA^2B,$$
but I can't conclude that it's $I$.
Assuming $A\in \mathbb{R}^{n\times n}$, let $f_A : \mathbb{R}^n \to \mathbb{R}^n, x\mapsto Ax$ be the associated linear map. Since $BA = I$, we have $f_B\circ f_A = \operatorname{id}$, so $f_A$ is injective. But then $f_A$, as a linear map between finite-dimensional vector spaces of same dimension, is also surjective, hence bijective and so $A$ is invertible.

This is because $n = \operatorname{dim}(\ker f_A) + \operatorname{dim}(\operatorname{im} f_A)$. Since $f_A$ is injective, $\operatorname{dim}(\ker f_A) = 0$, so $\operatorname{dim}(\operatorname{im} f_A) = n$, hence $f_A$ is surjective.
If you can try $B = \text{adj}(A)/\det(A)$ and show by brute force computation that $BA = I = AB$.

You would only need $B^jA_i = \delta_{ij}$ and $A^iB_j = \delta_{ij}$. (Superscript means row vector and subscript meaning column vector index, and $\delta_{ij}$ means indicator of $i = j$).  

You only need to think about the relationship between the definitions of $\text{adj}$ and $\det$. See https://en.wikipedia.org/wiki/Adjugate_matrix

If additionally you know how transpose relates to inverse, I think you only need $B^jA_i = \delta_{ij}$.