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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
$$\mathrm{tr}(\mathcal{E}_m(\rho) )= \sum_n \langle n | M_m \rho M_m^{\dagger} | n\rangle = \sum_n \langle n | M_m | \psi\rangle \langle \psi | M_m^{\dagger} | n\rangle $$
to $\langle\psi|M_m^{\dagger}M_m|\psi\rangle$?
Because after this I can’t find any idea to boil down to this.
This is because:
$$ \sum_n \langle n | M_m | \psi\rangle \langle \psi | M_m^{\dagger} | n\rangle = \sum_n \langle \psi | M_m^{\dagger} | n\rangle \langle n | M_m | \psi\rangle = \langle \psi | M_m^{\dagger} I M_m | \psi\rangle = \langle \psi | M_m^{\dagger} M_m | \psi\rangle $$
note that $\sum_n|n\rangle\langle n| = I $
