= Solution
We prove the stronger <finite branching bound in a complete market>: modulo null sets, $\mathcal F_t$ is generated by at most $n^t$ atoms. Here an atom is a positive-probability event which cannot be split into two positive-probability measurable pieces.
At time zero there is one atom because $\mathcal F_0$ is trivial. Suppose $\mathcal F_{t-1}$ has at most $n^{t-1}$ atoms, and fix one such atom $A$. Any <predictable process> holdings over $(t-1,t]$ are constant on $A$, so the restriction to $A$ of every time-$t$ replicable payoff lies in
$$
\operatorname{span}\{P_t^1|_A,\ldots,P_t^n|_A\},
$$
a <vector space> of <dimension of a vector space> at most $n$. If $A$ contained $n+1$ disjoint positive-probability $\mathcal F_t$ events, their indicators would have <linear independence> on $A$. <Market completeness> would replicate each indicator, contradicting that dimension bound.
For clarity, the absence of $n+1$ disjoint positive pieces implies that $A$ is a union of at most $n$ atoms: start with $A$ and repeatedly split any non-atom into two positive pieces. Each split increases the count by one. The process must stop before the count exceeds $n$, and at termination every piece is an atom. Thus each time-$(t-1)$ atom has at most $n$ successors. Induction gives at most $n^t$ atoms at time $t$.
Each disjoint positive-probability $\mathcal F_t$ event contains at least one distinct atom. Therefore
$$
\boxed{k\leq n^t.}
$$
The argument includes $t=0$ and uses the ability to replicate the indicator claims at every intermediate maturity.
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