There's a common convention in analysis that the letter $c$ denotes some constant, the value of which may change each time it appears. (That is, when one writes $A\le cB$ and then later writes $X\le cY$ the two $c$'s are not necessarily the same constant.)
The author is assuming first that the reader knows that $\log(n)-(1+1/2+\dots+1/n)$ is bounded. This is easy to see:
$$\begin{align}\log(n)-\log(n-1)-\frac1n&=\int_{n-1}^n\left(\frac 1t-\frac1n\right)\,dt
\\&=\int_{n-1}^n\frac{n-t}{nt}\,dt.\end{align}$$
Now $t\in[n-1,n]$ implies that $$0\le\frac{n-t}{nt}\le\frac1{n(n-1)},$$
so $$0\le \log(n)-\log(n-1)-\frac1n\le\frac1{n(n-1)}.$$Since $\sum\frac1{n(n-1)}<\infty$ it follows from the triangle inequality that there exists $c$ such that $$\left|\log(n)-(1+\frac12+\dots+\frac1n\right|\le c.$$
Hence if $N$ is fixed there exists $c_N$ such that$$\left|\log(n)-(\frac1N+\frac1{N+1}+\dots+\frac1n\right|\le c_N$$for $n>N$. So if $R>0$ and $n>N$ then $$e^{-R(\frac1N+\dots\frac1n)}\ge e^{-R(\log(n)+c_N)}
=e^{-Rc_N}e^{-R\log(n)}=ce^{-R\log(n)},$$where $c=e^{-Rc_N}$.
In case it comes up elsewhere, the same argument shows that$$e^{-R(\frac1N+\dots+\frac1n)}\le e^{Rc_N}e^{-R\log(n)}.$$
