# How was the following lower bound on ${n \choose pn}$ derived?

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.

Note that for any $t > 0$ $$\frac{-\tfrac12\ln (2\pi p(1-p)) - \tfrac12\ln(n) - \ln(1+o(1))}{n^t} \to 0$$ so the exponential expression satisfies $\exp(-o(n^t))$. So
$$\binom{n}{pn}\geq \exp(-o(n))\left(\frac{1}{p}\right)^{np} \left( \frac{1}{1-p}\right)^{n(1-p)}$$