Average ocupancy in an ideal gas at high-temperature In David Chandler's 'intro to statistical mechanics' he states that for an ideal gas at high-temperature
$$
\langle n_j\rangle=\langle N\rangle\frac{e^{-\beta \epsilon_j}}{\sum e^{-\beta \epsilon_j}}
$$
Which I can believe from intuition, but he losses me on the derivation. 
Starting with the general form for the occupancy of a boson or Fermion gas:
$$
\langle n_j \rangle=[e^{\beta (\epsilon_j-\mu)} \pm 1]^{-1}
$$
Then at high-temperature, $\beta \rightarrow 0$, [ page 101 (b) ]
$$
e^{\beta(\epsilon_j-\mu)}>>1
$$
So the assumption can be made that 
$$
\langle n_j \rangle=e^{-\beta (\epsilon_j-\mu)}
$$
Which would make sense if $e^{\beta(\epsilon_j-\mu)}>>1$, but it sure seems like that $ e^{\beta(\epsilon_j-\mu)}\approx 1$ in that case. Is there another assumption that's made here?
He does say:

Note that if this eqn is true for all $\epsilon_j$, then $-\beta \mu >>1 $

so, $\mu \rightarrow -\infty$ at high-temperature? That doesn't strike me as something that's obvious. 
The rest of the derivation:
$$
\langle N  \rangle = \sum \langle n_j \rangle =\sum e^{-\beta(\epsilon_j-\mu)}=e^{\beta \mu} \sum e^{-\beta \epsilon_j} \\
e^{\beta \mu} = \frac{\langle N \rangle}{\sum e^{-\beta \epsilon_j}}
$$
using $\langle n_j \rangle = e^{-\beta(\epsilon_j - \mu)}$
$$
\langle n_j\rangle=\langle N\rangle\frac{e^{-\beta \epsilon_j}}{\sum e^{-\beta \epsilon_j}}
$$
 A: There are indeed other assumptions in the derivation you quote. Namely, Chandler considers the classical limit of an ideal quantum mechanical gas with average particle number
$ <N> = \sum_j <n_j> = \sum [ e^{\beta (e_j - \mu)} ± 1 ]^{-1} $ 
(plus for Fermi-Dirac, minus for Bose-Einstein statistics). In the classical limit (low density) there are many more single particle states than particles. Thus $ <n_j>$ is much smaller than 1, which implies
$ e^{\beta ( \epsilon_j - \mu)} >> 1 \ \ ($ i.e. the "$± 1$" in the eqn above becomes irrelevant)
In other words, the distinction between FD and BE statistics vanishes in the classical limit.
The second part of your question concerns the chemical potential $\mu$, which is shown to be just the standard Boltzmann factor in the classical limit. You don't need to consider any limits on $\mu$, just the fact that $<n_i>  \ \rightarrow  \ \exp(-\beta(\epsilon_i - \mu))$ in the classical limit.
A: Actually, it does make sense that $\mu \rightarrow - \infty$
Given the ideal chemical potential for in ideal gas:
$$
\mu = -k_B T\ln \left( \frac{V}{N} \left(\frac{mk_B T}{2 \pi \hbar}\right)^{3/2} \right )
$$
so
$$
\mu \beta \sim - \ln(T) \\
\:\\
\therefore \lim_{T \rightarrow \infty} -\mu \beta >> 1
$$
