Just wanted some input to see if my proof is satisfactory or if it needs some cleaning up.
Here is what I have.
Proof:Suppose $A$ is square and invertible and for the sake of contradiction let $0$ be an eigenvalue. Consider, $(A-\lambda I)\cdot v = 0$ with $\lambda=0 $
$$\Rightarrow (A- 0\cdot I)v=0$$
We know $A$ is an invertible and in order for $Av = 0$, $v = 0$, but $v$ must be non-trivial such that $\det(A-\lambda I)=0$
Here lies our contradiction.
Hence, $0$ cannot be an eigenvalue.
Suppose $A$ is square and has an eigenvalue of $0$. For the sake of contradiction lets assume $A$ is invertible.
Consider, $Av = \lambda v$, with $\lambda = 0$ means there exists a non-zero $v$ such that $Av = 0$. This implies $Av = 0v \Rightarrow Av = 0$
For an invertible matrix $A$, $Av = 0$ implies $v = 0$ So, $Av = 0 = A\cdot 0$
Since $v$ cannot be $0$,this means $A$ must not have been one-to-one.
Hence, our contradiction, $A$ must not be invertible.