= Solution
Include all fixed resource states and all measurement records in an environment, so the entire physical protocol has an <linear isometry> $U$ from the unknown input to a larger pure output. If every recipient receives a perfect copy, its <reduced density matrix> is the pure projector $|\psi\rangle\langle\psi|$. A pure marginal forces factorization: write a joint <vector> in a <basis> beginning with $|\psi\rangle$; its squared components along every <orthogonal> <basis> <vector> vanish because their marginal <probabilities> are zero. Applying this successively to all recipients gives
$$
U(|\psi\rangle|R\rangle)=|\psi\rangle^{\otimes N}|E_\psi\rangle,
$$
where $|R\rangle$ is the fixed initial resource and $|E_\psi\rangle$ may depend on the input. This includes Alice's discarded systems and every possible outcome record.
Choose two distinct nonorthogonal <pure states>, for example $|0\rangle$ and $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$, and write their overlap as $s$. Preservation of the <inner product> by the <linear isometry> requires
$$
s=s^N\langle E_\psi|E_\varphi\rangle.
$$
Normalized environment <vectors> have overlap of absolute value at most one. Thus $0<|s|<1$ would imply $|s|\leq|s|^N$, impossible for $N>1$. Hence \b[perfect transmission of a copy to every recipient is impossible for $N>1$]. This proves the <no-cloning theorem> obstruction even though Alice keeps no copy herself.
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