= Solution
Put $z=e^{\beta\mu}$, with $\beta=1/(k_BT)$. For a single quantum state of <energy> $\epsilon$, dilute <Maxwell–Boltzmann statistics> gives $\ln\Xi_\epsilon\simeq ze^{-\beta\epsilon}$. For bosons this is the leading expansion of $-\ln(1-ze^{-\beta\epsilon})$; the occupation must be small for that expansion. Thus the <grand potential> is
$$
\boxed{\Phi(\epsilon)=-k_BT\ln\Xi_\epsilon\simeq-k_BT e^{-\beta(\epsilon-\mu)}.}
$$
With the ground-state <energy> set to zero, a dilute classical gas requires $e^{\beta\mu}\ll1$, giving negative $\mu$. This sign is relative to the chosen <energy> zero. For an ideal Bose gas, $\mu$ approaches the ground <energy> from below at condensation, so it approaches zero with this convention.
The printed degeneracy formula requires $n_x,n_y,n_z=0,1,2,\ldots$, rather than strictly positive integers. With zero-point <energy> omitted these are the appropriate oscillator quantum numbers. The <stars and bars> count of three nonnegative integers summing to $n$ gives
$$
\boxed{d_n=\binom{n+2}{2}=\frac{(n+2)!}{n!\,2!}.}
$$
There is one ground state at $n=0$. If the integers were strictly positive, the count would instead be $\binom{n-1}{2}$ for $n\geq3$, inconsistent with both the printed formula and its <energy> reference.
For large $n$, $d_n\simeq n^2/2$. Consecutive <energy> shells are separated by $\hbar\omega$, giving the continuum <density of states>
$$
\boxed{g(E)\simeq\frac{n^2}{2\hbar\omega}=\frac{E^2}{2(\hbar\omega)^3}.}
$$
Use this when $k_BT\gg\hbar\omega$, and integrate the single-state grand potential over <energy>:
$$
\Phi_{\rm tot}\simeq-\frac{k_BT e^{\beta\mu}}{2(\hbar\omega)^3}\int_0^\infty E^2e^{-\beta E}dE=-\frac{(k_BT)^4}{(\hbar\omega)^3}e^{\beta\mu},
$$
because the integral is $2/\beta^3$. Particle number follows by differentiating at fixed <temperature> and trap <frequency>:
$$
N=-\frac{\partial\Phi}{\partial\mu}=\left(\frac{k_BT}{\hbar\omega}\right)^3e^{\beta\mu},\qquad\boxed{\mu=k_BT\ln\left[N\left(\frac{\hbar\omega}{k_BT}\right)^3\right].}
$$
These are the high-temperature expressions for the <classical grand potential in a harmonic trap>.
Extrapolating to $\mu=0$ gives the requested transition scale
$$
\boxed{T_c\sim\frac{\hbar\omega}{k_B}N^{1/3}.}
$$
It is an order-one estimate because Maxwell-Boltzmann occupations cease to be accurate near condensation. The <Bose-Einstein distribution> gives the more precise ideal-trap result $N_{\rm ex}^{\max}=\zeta(3)(k_BT/\hbar\omega)^3$, and hence $T_c=(\hbar\omega/k_B)[N/\zeta(3)]^{1/3}$, about $0.94$ times the estimate. Excess particles populate the ground state after the excited-state capacity saturates: <Bose condensation in a three-dimensional harmonic trap>.
The <observation of Bose-Einstein condensation in rubidium in 1995> implemented laser precooling followed by radio-frequency evaporation from a magnetic trap. Imaging the released cloud revealed a narrow anisotropic component within a broader thermal population. This provides an experimental connection to the population-counting argument; the measured expansion also reflects interactions and imaging resolution, not just the ideal-trap <density of states>.<Observation of Bose-Einstein Condensation in a Dilute Atomic Vapor by M. H. Anderson and others>[1995, pp. 198–201, Figs. 1–4]{ref}
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