Evaluating a limit of the truncated exponential series motivated by the prime number theorem for $k$ distinct prime factors. If $\pi_k(n)$ is the cardinality of numbers with k factors (repetitions included) less than or equal n, the generalized Prime Number Theorem is:
$$\pi_k(n)\sim \frac{n}{\ln n} \frac{(\ln \ln n)^{k-1}}{(k-1)!}.$$
I noticed it appears true that
$$\lim_{n \to\infty}\ \sum_{k=1}^{n}\frac{2^n}{\ln 2^n} \frac{(\ln\ln 2^n)^{k-1}}{(k-1)!} = 2^n ,$$
which makes sense to me. In my attempts to prove this I could only get a few steps along.  How can this be done?

Edit: Looking through Ramanjuan's Collected Papers in the 32d paper I notice he has the following: 
$$[x] = \{\pi_1(x) + \pi_2(x)+\pi_3(x)...\}.......(1)$$ 
and 
$$x = \frac{x}{\ln x}\{1 + \ln\ln x + \frac{(\ln\ln x)^2}{2!}...\}....(2)$$
He says that (1) and (2) are "obvious." The second was not obvious to me, but can be found by letting y = $\ln\ln x$ and using the Taylor series for e. I'm including this for completeness because (1) above is the idea behind the original question. 
 A: Consider, for some constant $\kappa$,
$$ \begin{eqnarray}
    c_n(\kappa) &=& \frac{1}{n \kappa }\sum_{k=0}^{n-1} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} = \frac{1}{n \kappa } \left( \sum_{k=0}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} - \sum_{k=n}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} \right) \\ \
& =& 1 - \frac{1}{n \kappa } \sum_{k=n}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!}
\end{eqnarray}
$$
You are interested in the behavior of $2^n c_n(\ln 2)$ for large $n$.
The latter sum vanishes by the Stolz-Cesàro theorem. Indeed, for $b_n = n \kappa$, and $a_n = \sum \limits_{k=n}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!}$, 
$$
   \lim_{n \to \infty} \frac{a_{n+1} - a_{n}}{b_{n+1}-b_n} =  \lim_{n \to \infty}\left(-\frac1{\kappa} \cdot \frac{ \left(\ln\left( n \kappa\right)\right)^{n-1}}{(n-1)!} \right)= 0
$$
where the Stirling approximation can be used to prove the limit.
Thus, $\lim \limits_{n \to \infty} c_n(\kappa) = 1$.
