Show that for the eigenvector, $u_i$ of eigenvalue $0$ of the matrix $A A^{\dagger}$, then $\langle u_i | Ax \rangle=0$

Question

Consider the eigenvector $\mathbf{u_i}$ corresponding to the eigenvalue $\lambda_i = 0$ for the matrix $AA^{\dagger}$, for an arbitrary $M \times N$ matrix $A$. Hence:

$$AA^{\dagger}\mathbf{u_i} = 0$$

Show that if $\mathbf{b}$ is in the range of A (i.e. $A\mathbf{x}=\mathbf{b}$)

then:

$$\mathbf{u_i}^{\dagger} \mathbf{b} = 0$$

My attempt:

\begin{align}AA^\dagger \mathbf{u_i} &= 0 \\ \mathbf{u_i}^\dagger AA^\dagger &= 0 \tag{1} \\ \mathbf{u_i}^{\dagger} \mathbf{b} &= \mathbf{u_i}^\dagger A\mathbf{x} \tag{2}\end{align}

I am unsure how to link $(1)$ and $(2)$ because it is not given that A is unitary, so i cannot assume $A^{\dagger}A = I$

Any help is appreciated!