Showing a function goes to zero exponentially fast I'm trying to find an exponentially decaying upper bound for the function
$$
f(\mu)=\frac{\mu(\rho-1)e^{-\mu(\rho-1)a}}{\rho-e^{-\mu(\rho-1)a}},
$$
as $\mu\rightarrow\infty$ where $\rho>1$ and all variables are nonnegative, i.e., in the form 
$$
f(\mu)\leq C_{1}e^{-C_{2}\mu}.
$$
I can bound $f(\mu)$ as follows:
$$
f(\mu)=\frac{\mu(\rho-1)e^{-\mu(\rho-1)a}}{\rho-e^{-\mu(\rho-1)a}}\leq\frac{\mu(\rho-1)e^{-\mu(\rho-1)a}}{\rho-1}=\mu e^{-\mu(\rho-1)a}.
$$
I know that $\mu(\rho-1)e^{-\mu(\rho-1)a}$ is quasi-concave with
a maximum of $1/(ea(\rho-1))$ but I haven't been able to find a bound
with $\mu$ in the exponent as above.

Substituting $x = \mu (\rho-1)a$, you're looking for $C_1$ and $C_2$ such that
\begin{equation}
\frac{x e^{-x}}{\rho - e^{-x}} \le C_1 e^{-C_2 x}
\end{equation}
Taking the natural log of both sides and rearranging gives
\begin{equation}
\ln x - \ln\left[C_1(\rho - e^{-x})\right] \le (1 - C_2) x
\end{equation}
So as long as $C_2 < 1$, your function will be bounded above by $C_1 e^{-C_2 x}$ for sufficiently large $x$.
