= Solution
Assume a dilute classical <ideal gas> of $N$ indistinguishable, noninteracting particles, with translational states dense enough to replace their sum by a continuum integral. Separate translation from all internal degrees of freedom. Then
$$
Q_{\rm trans}=\frac{q_{\rm trans}^{\,N}}{N!},\qquad
q_{\rm trans}=\frac V{\Lambda^3},\qquad
\Lambda=\frac h{\sqrt{2\pi mk_{\rm B}T}}.
$$
Using <Stirling's approximation> in the <Helmholtz free energy>,
$$
A_{\rm trans}=-Nk_{\rm B}T\left[\ln\frac{V}{N\Lambda^3}+1\right].
$$
Differentiate at fixed $N,V$. Since $\Lambda^{-3}\propto T^{3/2}$,
$$
S_{\rm trans}=-\left(\frac{\partial A_{\rm trans}}{\partial T}\right)_{N,V}
=Nk_{\rm B}\left[\ln\frac{V}{N\Lambda^3}+\frac52\right].
$$
Only after this derivative set $N=N_A$ and $V=RT/p^\circ$ for one mole at <standard pressure>:
$$
\boxed{\frac{S^\circ_{\rm trans,m}}R
=\ln\left[\left(\frac{2\pi mk_{\rm B}T}{h^2}\right)^{3/2}\frac{RT}{p^\circ}\right]
-\ln N_A+\frac52.}
$$
This is the translational <Sackur-Tetrode equation>. The $N!$ factor is essential for the correct extensive <entropy>; replacing $V$ by a temperature-dependent standard molar volume before taking the fixed-volume derivative would give the wrong constant term.
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