The derivation of entropy of perfect gas is given as follows in my textbook:
Suppose we have a vessel containing a volume $V$ of a perfect gas. The gas has $N$ molecules and is in a state of equilibrium at a temperature $T$. As the gas is in equilibrium it is in the most probable macrostate.
Let the $N$ molecules be divided into $k$ compartments (energy intervals) numbered $(1,2,...,k)$ having $g_1,g_2,..,g_k$ equal sized cells and $n_1,n_2,...n_k$ molecules respectively. As a perfect gas obeys Maxwell Boltzmann statistics, the thermodynamic probability of the macrostate ($n_1,n_2,...,n_k$) is given by
$$W(n_1,n_2,..,n_k)=\frac{N!(g_1)^{n_1}(g_2)^{n_2}...(g_k)^{n_k}}{n_1!n_2!...n_k!}$$
or $$W=N!\prod_{i=1}^{i=k}\frac{(g_i)^{n_i}}{n_i}$$
where $$n_i=g_ie^{-\alpha}e^{-\beta u_i}=g_ie^{-\alpha}e^{-u_i/kT}$$
Taking natural logarithms on both sides we have
$$\ln(W)=\ln(N!)+\sum_{i=1}^{i=k}n_i\ln(g_i)-\sum_{i=1}^{i=k}\ln(n_i!)$$
Now using $\ln(N!)=N\ln(N)-N$ for $N\to\infty$;
$$\ln(W)=N\ln(N)-\sum_{i=1}^{i=k}n_i\ln(\frac{n_i}{g_i})$$
Substituting $n_i=g_ie^{-\alpha}e^{-\beta > u_i}=g_ie^{-\alpha}e^{-u_i/kT}$
$$\ln(W)=N\ln(N)-N\ln(e^{-a})+\sum_{i=1}^{i=k}\frac{n_iu_i}{kT}\ln(e)$$
But $\sum_{i=1}^{i=k}n_iu_i=U$ where $U$ is total energy of system
$$\therefore > \ln(W)=N\ln(N)-N\ln(e^{-a})+U/kT=N\ln(N/e^{-a})+U/kT=N\ln(N)e^{a}+U/kT$$
For a system obeying Maxwell Boltzmann statistics $$e^{-\alpha}=\frac{Nh^3}{V}(\frac{\beta}{2\pi m})^{3/2} \implies Ne^{\alpha}=V/h^3(2\pi mkT)^{3/2}$$
Substituting the above value of $Ne^{\alpha}$ in equation (iii) we have $$\ln(W)=N\ln[\frac{V}{h^3}(2\pi mkT)^{3/2}]+\frac{U}{kT}$$
For mono-atomic gas $U=\frac{3}{2}NkT$.
Now, using the statistical definition of entropy:
$$S=k\ln(W)=\frac{3}{2}+Nk\ln[\frac{V}{h^3}(2\pi mkT)^{3/2}]$$
Now my question is:
Why is $e^{-\alpha}=\frac{Nh^3}{V}(\frac{\beta}{2\pi m})^{3/2}$ true for Maxwell Boltzmann statistics and how is the equation derived?