Prove that if $n\times n$ matrix $A$ is nonsingular, then it is invertible. I know that nonsingularity means having unique soln. $x$ for every RHS b in the system $Ax= b$. But how do I use this definition to come up with a proof for the problem ? I really have tried but to no avail.

An even more direct proof without presuming knowledge of as much as determinants:
You can solve $A\vec{x}_1=[1,0,\dotsc,0]^T$ for $\vec{x}_1$. Consider the matrix formed by $\vec{x}_1$ then $n-1$ columns of zeros, call it $B$.
Then it is clear that $AB=\begin{bmatrix}1&0&\dotsc\\0&0&\dotsc\\\vdots&\vdots&\ddots\end{bmatrix}$ by the properties of matrix multiplication.
Next, solve for $A\vec{x}_2=[0,1,\dotsc,0]^T$, $A\vec{x}_3=[0,0,1,\dotsc,0]^T$ and so on.
Constructively, the matrix $A^{-1}=\begin{bmatrix}\vec{x}_1&\vec{x}_2&\vec{x}_3&\dotsc\end{bmatrix}$
satifies $AA^{-1}=I$.
