Prove that if $A \cdot A^t = I$ then $A$ is invertible. I have the following statement:

Prove or disprove that if $A \cdot A^t = I$ and $A$ is a square matrix then $A$ is invertible.

I know that if $A$ is invertible and $A^t = A^{-1} \implies A \cdot A^t = I$
Since $A^t = A^{-1} \implies A\cdot A^t = A\cdot A^{-1}\to A \cdot A^t =I$
But i can't prove the double implication, i.e $A$ is invertible and $A^t = A^{-1} \iff A \cdot A^t = I$
I tried to show that $\det A $ is always non zero, but i have not succeeded. Perhaps the algebraic approach is more viable. Any help is appreciated.
 A: If $AB=I$ then $1=\det(AB)=\det(A)\det(B)$ then    $\det(A) \neq 0$ and $A$ is invertible. In particular you can take $B=A^{T}$.
From this the equivalence is quite easy. I will leave that to you.
A: Suppose $AA^t=I$ and assume $A$ is not invertible. Then we have that $$1=\det(I)=\det(AA^t)=\det(A)\det(A^t)=0\cdot\det(A^t)=0$$ and we can conclude that $1=0$, but this is a contradiction, so our initial assumption that $A$ is not invertible is false, therefore $A$ must be invertible, and let the inverse of $A$ be $A^{-1}$.
Then we can conclude that $$A^{-1}AA^t=A^{-1}I\Longrightarrow A^t=A^{-1}$$ and
$$A^tA=A^{-1}A=I $$
A: We proceed by contradiction. Suppose there exists $v \neq 0$ such that $Av = 0$. Then $\lambda = 0$ is an eigenvalue of $A$. Since $A$ and $A^{t}$ are similar, their eigenvalues coincide. Hence, $\lambda = 0$ is also an eigenvalue of $A^{t}$. There exists $u \neq 0$ such that $A^{t}u = 0$. But we get a contradiction since $u = Iu = AA^{t}u = 0$.
A: If $A$ is real the conclusion is immediate because  $A$ and $A^TA$ have the same nullspace.
A: Here is an approach that doesn't use determinants.  It is slightly longer, but gets more to the heart of what is going on, which is that the rank-nullity theorem shows that square matrices are invertible if and only if they are injective, if and only if they are surjective.  This is similar to how a set map from $\{1,2,\ldots, n\}$ to itself is invertible if and only if it is injective, if and only if it is surjective.
If we can find a $B$ such that $AB=I$, then for every vector $v$, we can find $w$ such that $Aw=v$ by setting $w=Bv$.  In fact, having a right inverse is equivalent to being surjective.  If an $n\times n$ matrix is surjective, it is of rank $n$, and by the rank-nullity theorem, it must have kernel $0$, which means it is injective.  Since $A$ is injective and surjective, it is bijective and therefore invertible.  Let us show that $B$ is that inverse.  $$B=IB=(A^{-1}A)B=A^{-1}(AB)=A^{-1}I=A^{-1}.$$
