= Solution
Take a riskless gross return $R>0$ per period and a <stock> whose independent multiplicative moves are $u$ and $d$, with physical up probability $p\in(0,1)$. Assume $0<d<R<u$. With $K_k$ up moves by date $k$,
$$
B_k=R^k,\qquad S_k=S_0u^{K_k}d^{k-K_k}.
$$
A <self-financing portfolio> chooses each <stock> holding from information available before the next move, and invests the remainder in the <bank account>. At any node the two possible next payoffs can be matched by two holdings because $u\ne d$. Backward induction therefore makes the <binomial market> complete.
The unique <risk-neutral probability> of an up move is determined by the discounted-stock <martingale> equation:
$$
qu+(1-q)d=R,\qquad\boxed{q=\frac{R-d}{u-d}\in(0,1).}
$$
The <risk-neutral measure> assigns product probabilities to the successive moves. A path with $K=K_n$ up moves has physical probability $p^K(1-p)^{n-K}$ and pricing probability $q^K(1-q)^{n-K}$. Its <Radon-Nikodym derivative> is therefore
$$
Z=\frac{dQ}{dP}=\left(\frac qp\right)^K\left(\frac{1-q}{1-p}\right)^{n-K}.
$$
Since $S_n=S_0d^n(u/d)^K$, eliminate $K$ to obtain the requested terminal-price expression:
$$
\boxed{Z=\left(\frac{1-q}{1-p}\right)^n\left(\frac{S_n}{S_0d^n}\right)^\beta,\qquad\beta=\frac{\log\!\left(q(1-p)/(p(1-q))\right)}{\log(u/d)}.}
$$
The pricing density depends only on the terminal <stock> price even though different paths to that price have different intermediate histories.
For $w_0>0$, a nonnegative terminal wealth $X$ with initial cost $w_0$ satisfies
$$
\mathbb E_P[ZX]=\mathbb E_QX=R^nw_0.
$$
The <utility function> has derivative $v'(x)=x^{-(\gamma-1)/\gamma}$. Put $\eta=\gamma/(\gamma-1)>1$. Its first-order condition makes $v'(X)$ proportional to $Z$, so $X=CZ^{-\eta}$. Choosing $C$ to satisfy the pricing <budget constraint> gives the <binomial power-utility terminal wealth>
$$
\boxed{X_*=\frac{R^nw_0}{\mathbb E_P[Z^{1-\eta}]}Z^{-\eta}.}
$$
All states have positive probability and the state space is finite, so this payoff is positive, integrable and attainable. The normalizing constant is explicit: independence gives
$$
\mathbb E_P[Z^{1-\eta}]=\left[p\left(\frac qp\right)^{1-\eta}+(1-p)\left(\frac{1-q}{1-p}\right)^{1-\eta}\right]^n.
$$
Combining these two formulas with the terminal-price expression for $Z$ gives $X_*$ entirely as a function of $S_n$.
To prove global optimality, <strict concavity> gives $v(X)-v(X_*)\le v'(X_*)(X-X_*)$. Because $v'(X_*)=C^{-1/\eta}Z$, the <expectation> of the right side is zero for any other payoff satisfying the same pricing <budget constraint>. Equality requires $X=X_*$ almost surely. Backward replication supplies its <self-financing strategy>, so this static optimum is attainable by trading. If $w_0=0$ and terminal wealth must be nonnegative, the only feasible terminal wealth is zero. For $p=q$, the density is one and the optimizer is the riskless wealth $R^nw_0$.
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