= Solution
An $e^k$-runner abacus separates bead positions by their residue modulo $e^k$. Write such a residue in base $e$ as
$$
a_0+a_1e+\cdots+a_{k-1}e^{k-1}.
$$
Taking one $e$-quotient sorts beads by $a_0$ and divides their positions by $e$; applying the operation again sorts by $a_1$, and so on. After $k$ stages, the iterated construction has selected exactly the same $e^k$ residue classes as the single $e^k$-quotient. The two conventional orderings may list the base-$e$ digits in opposite order, producing only a permutation of components.
Equivalently, induction on $k$ applies the same argument to every component of $Q_e(\lambda)$ and identifies the resulting $e^{k+1}$ runner partitions. Hence the <iterated quotient equals a power quotient up to permutation> statement is
$$
\boxed{Q_{e^k}(\lambda)\text{ is a permutation of }TQ_e(\lambda)_k}.
$$
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