Suppose $A$ is invertible matrix, we want to prove that in the canonical form that matches $A$ there is no zero rows.
So I proved the following that if the canonical form that matches there is a row of zeros then $A$ isn't invertible matrix then in something along that, let $\hat{A}$ be the canonical form of A. Since $\hat{A}$ have at least one row of zeros, there is less then $n$ rows with leading '1', so there is at least one free variable. this is the solution dimension of the homogeneous equations. Hence there are infinite solutions.
Suppose we have $T_{A}\colon\mathbb{F}^{n}\to\mathbb{F}^{n}$, linear transformation defined by multiplying by $A$, then $\ker T_{A}=\operatorname{null}\left(A\right)$
So $T_{A}$ isn't invertible, then $A$ isn't invertible, cause $A$ is invertible if and only if $T_{A}$ is invertible.
Those are just some guidlines I thought of to prove it,also forgive me if the terms are translated bad.
But I'm interested to know, if proving it straightforward $A$ invertible $\Rightarrow$ canonical form have no zero rows in it. Could be easier to prove? Or how would you do it?