While you can't _say_ that $\log(C_n)\to nK$ and $C_n \to e^{nK}$, your intuition is right -- you can make it precise.
The idea behind your assertions makes sense. If $\log(C_n)/n$ approaches $K$, then for large $n$, we should have $\log(C_n)$ approximately equal to $nK$. Thus we should have $\log(C_n)$ approximately equal to $e^{nK}$.
More precisely, what does this mean? It means that there is a function $\epsilon(n)$ such that $\epsilon(n)/n \to 0$ and $$ K - \epsilon(n)/n \leq \log(C_n)/n \leq K + \epsilon(n)/n. $$ Now we can do algebra all day long. $$ Kn - \epsilon(n) \leq \log(C_n) \leq nK + \epsilon(n) $$ and exponentiating (which is monotone) $$ e^{Kn}e^{-\epsilon(n)} \leq C_n \leq e^{Kn}e^{\epsilon(n)} $$
This actually gives more information than you asked for. It tells you not just that $C_n$ behaves like $e^{nK}$ for large $n$, it tells you how quickly $C_n$ tends to behave like that.