= Solution
On a probability <measure-preserving system>, for a finite <measurable partition> $\xi=\{A_1,\ldots,A_r\}$, the <entropy of a finite measurable partition> is
$$
H_\mu(\xi)=-\sum_{i=1}^r\mu(A_i)\log\mu(A_i),\qquad0\log0=0.
$$
Use natural logarithms, so <information entropy> is measured in nats; another fixed logarithm base rescales all answers. The <join of measurable partitions> is their common refinement, and write $\xi_a^b=\bigvee_{j=a}^bT^{-j}\xi$, with an empty join the trivial partition. The <entropy rate of a measurable partition> and <Kolmogorov-Sinai entropy> are respectively
$$
\boxed{h_\mu(T,\xi)=\lim_{N\to\infty}\frac1NH_\mu(\xi_0^{N-1}),\qquad h_\mu(T)=\sup_{\xi\text{ finite}}h_\mu(T,\xi)}.
$$
The block entropies form a <subadditive sequence>, so the first limit exists by the <Fekete lemma>. For finite partitions the <conditional entropy of finite measurable partitions> is $H(\eta\mid\zeta)=H(\eta\vee\zeta)-H(\zeta)$.
Put $c_0=H(\xi)$ and $c_k=H(\xi\mid\xi_1^k)$ for $k\ge1$. The <chain rule for information entropy>, applied from the last coordinate backwards, and measure preservation give
$$
H(\xi_0^{N-1})=\sum_{j=0}^{N-1}H(T^{-j}\xi\mid\xi_{j+1}^{N-1})=\sum_{k=0}^{N-1}c_k.
$$
The second equality uses invariance of the joint <partition atom> probabilities under the common pullback $T^{-j}$; invertibility is unnecessary. Since <conditioning reduces entropy>, $c_k$ decreases to a nonnegative limit $c$. The <Cesaro convergence of a sequence> of this convergent sequence has the same limit. Therefore
$$
\boxed{h_\mu(T,\xi)=\lim_{k\to\infty}H_\mu(\xi\mid\xi_1^k)}.
$$
Equivalently $h_\mu(T,\xi)=H_\mu(\xi\mid\mathcal F_1)$, where $\mathcal F_1=\sigma(\bigvee_{j\ge1}T^{-j}\xi)$. Here <conditional entropy of a countable measurable partition> conditioned on a <sigma-algebra> is computed using conditional <partition atom> probabilities; the <Martingale convergence theorem> gives continuity under increasing conditioning <sigma-algebras>. This is the <infinite-future formula for partition entropy rate>.
The <Kolmogorov-Sinai generator theorem> states that if a finite or countable <measurable partition> $\eta$ has finite <entropy of a countable measurable partition> and its iterates generate the whole completed <sigma-algebra> modulo null sets, then $h_\mu(T)=h_\mu(T,\eta)$. For an invertible system, generating means $\sigma(\bigvee_{j\in\mathbb Z}T^{-j}\eta)=\mathcal B$ modulo null sets. For a noninvertible system a <one-sided generator>, using $j\ge0$, suffices. The two-sided and one-sided versions must not be confused.
For a <Bernoulli shift> with discrete symbol probabilities $(p_i)$, the coordinate-zero <measurable partition> has independent coordinate iterates. Thus
$$
H(\eta_0^{N-1})=N\left(-\sum_i p_i\log p_i\right),\qquad\boxed{h_\mu(T)=-\sum_i p_i\log p_i}.
$$
For a finite alphabet the coordinate partition is a generator of finite <entropy of a finite measurable partition>; on the two-sided sequence space use all integer coordinate iterates, and on the one-sided space use the nonnegative ones. The <Kolmogorov-Sinai generator theorem> proves the displayed answer in both cases. The same calculation applies to countably many symbols when their <Shannon entropy> is finite, using the countable finite-entropy version of the theorem. If the <Shannon entropy> is infinite, merge all but the first $r$ symbols into one cell. These finite coordinate partitions have entropy rate $-\sum_{i\le r}p_i\log p_i-p_{>r}\log p_{>r}\to\infty$, so the system entropy is infinite. Zero-probability symbols contribute zero. In particular a fair $r$-symbol shift has entropy $\log r$.
For the final assertion, let $\mathcal F_0=\sigma(\bigvee_{j\ge0}T^{-j}\xi)$ and complete it modulo null sets. The approximation property forces $\mathcal F_0=\mathcal B$ modulo null sets. Indeed for each $A\in\mathcal B$, choose approximating sets from finite blocks with error tending to zero. Their indicators approach $\mathbf1_A$ in $L^2$, and $L^2(\mathcal F_0)$ is a closed <vector subspace>, so $A$ is $\mathcal F_0$-measurable modulo a null set.
Invertibility now gives $\mathcal F_1=T^{-1}\mathcal F_0=T^{-1}\mathcal B=\mathcal B$ modulo null sets. In particular the present partition $\xi$ is measurable with respect to its entire future, so $H(\xi\mid\mathcal F_1)=0$. The infinite-future formula gives $h_\mu(T,\xi)=0$. Since the given <one-sided generator> is also a two-sided generator, the <Kolmogorov-Sinai generator theorem> finishes the proof:
$$
\boxed{h_\mu(T)=0}.
$$
This is <finite one-sided generator of an invertible system forces zero entropy>. Invertibility is essential: a fair binary one-sided <Bernoulli shift> has a finite <one-sided generator> and <Kolmogorov-Sinai entropy> $\log2$.
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