In the middle of page 3 of this lecture on coding theory (linked), the following lower bound for ${n \choose pn}$ is derived, with $n$ and $q$ integers satisfying $n > 0$, $q \geq 2$, and $p \in [0, 1-1/q]$. Quoting:
By Stirling's formula, $m! = \sqrt{2\pi m}(m/e)^m(1 + o(1))$, [so] it follows that
${n \choose pn} \geq \left(\frac{1}{p}\right)^{pn} \left( \frac{1}{1-p} \right)^{(1-p)n} \exp(-o(n))$
My question is how exactly was this lower bound derived? In particular how does the first RHS factor $\left(\frac{1}{p}\right)^{pn}$ relate to ${n \choose pn}$; how does the second RHS factor $\left( \frac{1}{1-p} \right)^{(1-p)n}$ relate to ${n \choose pn}$; and finally how does $\exp(-o(n))$ relate to ${n \choose pn}$? I'm guessing that $\exp(-o(n))$ is a lower bound on $n!$ derived from the given Stirling formula, but I'm not entirely sure how it's obtained.