Why do ATA and AAT have the same eigenvalues?

Why is it that $A^TA$ and $AA^T$ have the same non-zero eigenvalues? A symbolic proof is not hard to find, but as usual, I prefer to find a way to visualize it in order to gain a better mathematical intuition.

Let $\vec{x}$ be an eigenvector of $A^TA$.

We start with vector $\vec{x}$. $A$ transforms $\vec{x}$ into some arbitrary vector $A\vec{x}$. This is multiplied by $A^T$ resulting in $A^TA\vec{x}$. But remember, we defined $\vec{x}$ as an eigenvector of $A^TA$, so by definition $A^TA\vec{x} = \lambda \vec{x}$.

Now we're almost back to where we started, except $\vec{x}$ is being multiplied by a scalar! So if $\lambda \vec{x}$ undergoes another linear transformation, the result will be $\lambda$ times the transformation of $\vec{x}$.

So what if we choose to multiply $\lambda \vec{x}$ by $A$? We get $\lambda A \vec{x}$. But to get to this point, we multiplied $A \vec{x}$ by $AA^T$.

This means that $A \vec{x}$ is an eigenvector of $A^TA$ with eigenvalue $\lambda$! This is the same eigenvalue that we found by multiplying $\vec{x}$ by $A^TA$!

Update: This is actually true for any matrices $AB$ and $BA$, not only a matrix and its transpose. Thanks reddit user etzpcm for pointing this out!