= Solution
The PDF's middle equality is \b[$(\omega^2)^M=(\omega^2)^{M[G]}$], concerning ordinal exponentiation. The TeX incorrectly turns the exponent into a subscript; preserving $\omega_2$ would be incompatible with forcing CH over many ground models.
Let $A=(\mathcal P(\omega))^M$ and use the <countable-condition collapse> of $A$ onto $\omega_1$:
$$
\boxed{\mathbb P=\{p:p\text{ is a countable partial function }\omega_1\to A\},}
$$
ordered by extension. Countable unions give lower bounds, so <closed forcing adds no short ordinal sequences>; in particular it adds no reals and preserves $\omega_1$. Dense requirements ensure that the union of the <generic filter> is a total surjection $g:\omega_1\to A$. Thus
$$
\boxed{\aleph_1^M=\aleph_1^{M[G]},\qquad
\mathcal P(\omega)^{M[G]}=\mathcal P(\omega)^M,\qquad
\operatorname{ran}(g)=\mathcal P(\omega)^{M[G]}.}
$$
Finally <forcing preserves ordinals>, and the recursion defining ordinal multiplication and exponentiation is absolute. Both models compute $\omega^2$ as the same order type of $\omega$ successive copies of $\omega$. This proves the printed middle equality too. Higher cardinals may collapse, which is allowed by the actual PDF.
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