Why does $sum_n langle n|M_mrho M_m^dagger|nrangle$ simplify to $langle psi|M_m^dagger M_m|psirangle$?

I was trying to derive the formula for $p(m)$ in exercise 8.2 on page 357 in Nielsen & Chuang. But I am wondering what rule I can apply to simplify this

$$mathrm{tr}(mathcal{E}_m(rho) )= sum_n langle n | M_m rho M_m^{dagger} | nrangle = sum_n langle n | M_m | psirangle
langle psi | M_m^{dagger} | nrangle $$

to $langlepsi|M_m^{dagger}M_m|psirangle$?

Because after this I can’t find any idea to boil down to this.