For $m \times n$ matrix $A$ with $m \ge n$, show ${\lVert A^+ \rVert}_2 = 1/\sigma_n$ where $A^+ = (A^*A)^{-1} A^*$

Question

For $m \times n$ matrix $A$ with $m \ge n$, show that the norm of the pseudoinverse ${\lVert A^+ \rVert}_2 = 1/\sigma_n$ where $A^+ = (A^*A)^{-1} A^*$ and $\sigma_n$ is the nth singular value of $A$.

$A^+ A = I_n$ so ${\lVert A^+ A \rVert} = 1 \le {\lVert A^+ \rVert} {\lVert A \rVert}$, so ${\lVert A^+ \rVert} \ge 1/\sigma_1$ which isn't helpful.

I also see that ${\lVert A^* A \rVert} = {\lVert A \rVert}^2 = \sigma_1^2$. I'm stuck on what to try next.