Proof that columns of an invertible matrix are linearly independent 
Explain why   the columns of  an $n \times n$ matrix $A$ are linearly
  independent when $A$ is  invertible.

The proof that I thought of was:
If $A$ is invertible, then $A \sim I$ ($A$ is row equivalent to the identity matrix). Therefore, $A$ has $n$ pivots, one in each column, which means that the columns of $A$ are linearly independent.
The proof that was provided was:
Suppose $A$ is invertible. Therefore the equation $Ax = 0$ has only one solution, namely, the zero solution. This means that the columns of $A$ are linearly independent.
I am not sure whether or not my proof is correct. If it is, would there be a reason to prefer one proof over the other?

As seen in the Wikipedia article and in Linear Algebra and Its Applications, $\sim$ indicates row equivalence between matrices.

I’ll use rows instead of columns, but the argument for columns holds similarly. Suppose a square matrix A is invertible that means it is row equivalent to the identity matrix. Now, observing that the rows of the identity are linearly independent, you can reapply the reverse operations on the rows of the identity to get the rows of A, this shows that the rows of A are linearly independent.
Suppose the rows of A are linearly independent. Applying row operations on the rows of A until you have a reduced echelon matrix, will give you a matrix with no zero rows and hence it must be the identity. Therefore A is row equivalent to the identity and so A is invertible.
