Prove if $x$ orthogonal to null space of $A$, $A^TAx = 0$ only when $x = 0$

I am learning linear algebra now, and encounter this problem:

If $x \in Nul(A)^\perp$ then $A^TAx = 0$ only if $x = 0$.

I think I solved this but just want to make sure I make every step correct:

Prove by contradiction: assume $x \neq 0$ but $A^TAx$. That is, $Ax = 0$.

By orthogonal decomposition $\mathbb{R}^n = R(A^T)\oplus Nul(A)$, since $x \in Nul(A)^\perp = R(A^T)$, it means $x \notin Nul(A)$, then if $x \neq 0$, $Ax \neq 0$, a contradiction.

Is my reasoning correct? Thank you!

The general idea is good. But you can clean it up a little, so it's easier to follow the logic.

For instance, there is no reason to go for contradiction here. Just start with $x\in Nul(A)^\perp$ and $A^TAx = 0$. See that $Ax = 0$. Then note that this means, by definition, that $x\in Nul(A)$.

But since we also have $x\in Nul(A)^\perp$, this means $x$ is perpendicular to itself. Conclude that $x$ must be $0$.