Recently, I was able to prove on my own using the Prime Number Theorem and integration, that the number of semiprimes (product of two primes) till $n$, say $\pi_2(n)$, satisfies the asymptotic expression: $$\pi_2(n) \sim \frac{n \ln{\ln{n}}}{\ln{n}}$$
After searching for similar results online, I found that if the number of numbers $<n$ which are the product of $k$ primes is defined as $\pi_k(n)$, then: $$\pi_k(n) \sim \frac{n(\ln{\ln{n}})^{k-1}}{\ln{n} (k-1)!}$$
But I also saw that these asymptotic expressions are poor approximations for small $n$ and that these approximations become better very slowly. I want to ask whether the following is true or not: $$\lim_{k \to \infty} \bigg(\frac{e^{2^k}(\ln{\ln{e^{2^k}}})^{k-1}}{\ln{e^{2^k}} (k-1)!} \cdot \frac{1}{\pi_k(e^{2^k})}\bigg)=1$$
I assume this will be true since the function $f(k)=e^{2^k}$ grows really fast, satisfying: $$f(k+1)=f(k)^2$$ and I want to know whether this growth rate is fast enough to make up for the slow change in accuracy of approximation that $\pi_k(n)$ gives. How do I prove this?
In general, what should the order of a function $g(k)$ be, so that: $$\lim_{k \to \infty} \bigg(\frac{g(k)(\ln{\ln{g(k)}})^{k-1}}{\ln{g(k)} (k-1)!} \cdot \frac{1}{\pi_k(g(k))}\bigg)=1$$
Any help is appreciated. Thanks in advance!
EDIT: It turns out that this isn't true. We have: $$\pi_k(n) = \frac{n(\ln{\ln{n}})^{k-1}}{\ln{n} (k-1)!}+O\bigg(\frac{n(\ln{\ln{n}})^{k-2}}{\ln{n}}\bigg)$$
The error term is really large and greater than the main term when $n=e^{2^k}$, disproving the claim. I need: $$g(x)>O\bigg(e^{e^{(k-1)!}}\bigg)$$ for the result to hold true.
Is there any result that adds more terms to the asymptotic expression for $\pi_k(n)$ such that the error term becomes smaller? (Preferably small enough so that the asymptotic expression will be a very good approximation for numbers of the order $e^{2^k}$.)