How is $ \left(1-\frac{p^2}{2mE}\right)^{3N/2-2} =\; \exp\left(-\frac{3N}{2}\frac{p^2}{2mE}\right)\;?$ How is 
$$ \left(1-\frac{p^2}{2mE}\right)^{3N/2-2} = \exp\left(-\frac{3N}{2}\frac{p^2}{2mE}\right)$$
(Karder, Statistical Physics of Particles, Page 107)
in the large $E$ limit. Here $N$ is particle, of the order of $10^{23}$, $E$ is the total energy.
I roughly guess that it should be $\exp(-\frac{p^2}{2m})$ since both $N$ and $E$ can be treated as infinitely large.

Update: a hint to solution is provided in the comments.
 A: You can use the approximations
$$1+x \simeq e^x$$ and 
$$(1+rx)  \simeq (1+x)^r$$
You can obtain 
$$(1+(-p^2/2mE))^{3N/2-2}$$ which can be approximated in the $N\gg1$ limit as $$(1+(-p^2/2mE))^{3N/2}$$ which is approximately equal to$$(1+\frac{3N}{2}(-p^2/2mE))\rightarrow \exp [1 + (-p^2/2mE)]^\frac{3N}{2}$$ by using the first foruma above. Here $\frac{3N}{2}$ is considered as a constant number much greater than 1, probably the ${3N/2-2}$ is the correction. 
A: Add a comment needs 50 reputation, and I got only 46 now. So I write my opinion here.
I have read the textbook, the original formula is
$$p(\vec{p_1})=(1-{{\vec{p_1}^2}\over {2mE}})^{3N/2-2}\cdots\cdots$$
So $\vec{p_1}$ is the momentum of only one particle in the ensemble. Considering the system has very large $N$, that is only a tiny proportion of total $E$, which makes the ${{\vec{p_1}^2}\over {2mE}}$ term approaches 0.
Then with the above comments of other guys, you can get the results. Here I think $3N/2$ makes no difference with $3N/2-2$ because $N$ is large
