= Solution
Write the spectral decomposition of the one-qubit <density operator> as $\pi=\sum_{j=0}^1\lambda_j|e_j\rangle\langle e_j|$. For a word $\mathbf j=(j_1,\ldots,j_n)$, the <tensor product> vector
$$
|e_{\mathbf j}\rangle=|e_{j_1}\rangle\otimes\cdots\otimes|e_{j_n}\rangle
$$
is an eigenvector of $\pi^{\otimes n}$, with eigenvalue
$$
\boxed{\lambda_{\mathbf j}=\lambda_{j_1}\cdots\lambda_{j_n}.}
$$
These product vectors form an orthonormal <eigenbasis>, allowing arbitrary orthonormal choices within degeneracies. Using $\sum_j\lambda_j=1$ and $0\log0=0$,
$$
\begin{aligned}
S(\pi^{\otimes n})
&=-\sum_{j_1,\ldots,j_n}\left(\prod_{s=1}^n\lambda_{j_s}\right)\sum_{s=1}^n\log_2\lambda_{j_s}\\
&=n\left(-\sum_j\lambda_j\log_2\lambda_j\right).
\end{aligned}
$$
Thus the <Von Neumann entropy> is
$$
\boxed{S(\rho^{(n)})=nS(\pi).}
$$
This calculation uses the spectral ensemble of the <memoryless quantum information source>; the originally emitted vectors need not equal these eigenvectors.
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