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For measurement, we know
$$\langle M \rangle = \sum_j p_j \langle \psi_j|M|\psi_j \rangle = \sum_j p_j \operatorname{tr}\left(|\psi_j \rangle \langle \psi_j|M\right).$$
My question is, how can we go from the first expression to the second one with trace operation? What's the math behind it?
First, note that by definition: $$Tr(X) = \sum_i X_{ii} = \sum_i \langle i| X | i\rangle $$ then now let $Y = |\psi\rangle \langle\psi| $ then this implies that
$$Tr(YX) = Tr(XY) = \sum_i \langle i|XY|i\rangle = \sum_i \langle i|X |\psi\rangle \langle\psi |i\rangle = \sum_i \langle \psi| i \rangle \langle i|X |\psi\rangle = \langle\psi|X|\psi\rangle = \langle X \rangle $$
I think the easiest way to see this is through the fact that the trace is independent of the basis in which we take it. That is, for any operator $A$ and any orthonormal basis $\{|i\rangle\}_i$ we can write $$ \mathrm{Tr}[A] = \sum_i \langle i | A | i \rangle. $$
So if $|\psi_j\rangle$ is normalised we can just choose any orthonormal basis which includes $|\psi_j\rangle$, let's write such a basis as $\{|\psi_j\rangle, |\psi_{j,1}^{\perp}\rangle, |\psi_{j,2}^{\perp}\rangle, \dots\}$. Then we have $$ \begin{aligned} \mathrm{Tr}[|\psi_j\rangle \langle\psi_j|M] &= \langle\psi_j|\psi_j\rangle\langle\psi_j|M|\psi_j\rangle + \sum_{i} \langle\psi_{j,i}^{\perp}|\psi_j\rangle\langle\psi_j|M|\psi_{j,i}^{\perp}\rangle \\ &= \langle\psi_j|\psi_j\rangle\langle\psi_j|M|\psi_j\rangle \\ &= \langle\psi_j|M|\psi_j\rangle, \end{aligned} $$ where on the second line we used the fact that $\langle\psi_{j,i}^{\perp}|\psi_j\rangle = 0$ for all $i$ as they form an orthonormal basis and on the final line we used normalization $\langle\psi_j|\psi_j\rangle = 1$.
$$\langle M\rangle=\sum_{j}p_{j}\langle\psi_{j}|M|\psi_{j}\rangle=\sum_{j}p_{j}\langle\psi_{j}|M\sum_{i}|e_{j}\rangle\langle e_{j}|\psi_{j}\rangle=\sum_{j}p_{j}\sum_{i}\langle e_{j}|\psi_{j}\rangle\langle\psi_{j}|M|e_{j}\rangle=\sum_{j}p_{j} tr(|\psi_{j}\rangle\langle\psi_{j}|M)$$
This involves two steps. In the first, we use the fact that the trace of a number, thought of as a $1 \times 1$ martix, is that number. In the second, we use the cyclic property of the trace.
Explicitly, the steps are
$$ \langle M \rangle = \sum_j p_j \langle \psi_j|M|\psi_j \rangle = \sum_j p_j \operatorname{tr}\left(\langle \psi_j|M|\psi_j \rangle\right) = \sum_j p_j \operatorname{tr}\left(|\psi_j \rangle \langle \psi_j|M\right). $$
