= Solution
Work internally in the ground model and write $P=\operatorname{Levy}(\aleph_0,\lambda)$. Conditions are finite partial <functions> with the coordinate-wise value bounds in the PDF, ordered by inclusion, so compatible conditions agree on their common domain and their union is a common strengthening.
Suppose there were $\lambda$ many pairwise incompatible conditions. Each fixed finite domain carries fewer than $\lambda$ possible <functions>: it uses finitely many <ordinals> below $\lambda$, and each coordinate has fewer than $\lambda$ choices. Regularity therefore lets us select $\lambda$ distinct domains. The finite-set form of the <generalized delta-system lemma> gives a subfamily of size $\lambda$ with common intersection $r$. There are fewer than $\lambda$ possible value assignments to the finite root $r$. Regularity gives a further subfamily of size $\lambda$ agreeing on the root. Any two of its conditions have a union in $P$, contradicting incompatibility. Thus \b[$P$ has the $\lambda$-chain condition]. Strong inaccessibility is more than is needed for this finite-support argument; regular uncountability suffices.
For each $0<\alpha<\lambda$ and $n<\omega$, requiring $(n,\alpha)$ to be in the domain is dense. For each $\xi<\alpha$, requiring $\xi$ to occur as a value at some $(n,\alpha)$ is also dense: choose a fresh $n$ and extend the condition. Hence the generic union defines, for each such $\alpha$, a <surjection> $g_\alpha:\omega\to\alpha$. Every ground <ordinal> below $\lambda$ is therefore countable in $M[G]$.
By the chain condition and the cardinal-preservation argument, $\lambda$ remains an uncountable <cardinal> in the extension; the same small-value argument preserves its regularity as well. All <ordinals> are unchanged. Since every smaller <ordinal> is countable and $\lambda$ is not,
$$
\boxed{\aleph_1^{M[G]}=\lambda.}
$$
This is the <finite Lévy collapse to omega-one>. The dense-set construction proves the collapse below $\lambda$, while the chain condition is what prevents collapsing $\lambda$ itself.
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